1. When we reverse the digits of the number 13, the increases by 18. How many other two digit numbers increases by 18 when their digits are reversed?
a) 5
b) 6
c) 7
d) 8
Discussion
Explanation: Let the numbers are in the form of (10x + y), so when the digits of the number are reversed the number becomes (10y + x)
According to question,
(10y + x) - (10x + y) = 18
9(y - x) = 18
→ y - x = 2
So, the possible pairs of (x, y) are (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9)
we need the number other than 13.
There are 6 possible numbers i.e. 24, 35, 46, 57, 68, 79
So, total numbers of possible numbers are 6
2. Find the LCM and HCF of 2.5, 0.5 and 0.175.
a) 17.5
b) 5
c) 7.5
d) 2.5
Discussion
Explanation:
$$\eqalign{ & 2.5 = \frac{{25}}{{10}}, \cr & 0.5 = \frac{5}{{10}}, \cr & 0.175 = \frac{{175}}{{1000}}, \cr} $$
Now,
LCM of two or more fractions is given by:
$$\eqalign{ & \frac{{{\text{LCM}}\,{\text{of}}\,{\text{Numerators}}}}{{{\text{HCF}}\,{\text{of}}\,{\text{Denominators}}}} \cr & \frac{{{\text{LCM}}\,{\text{of}}\,25,\,5,\,175}}{{{\text{HCF}}\,{\text{of}}\,10,\,10,\,1000}} \cr & = \frac{{175}}{{10}} \cr & = 17.5 \cr} $$
3. If A381 is divisible by 11, find the value of the smallest natural number A?
a) 9
b) 7
c) 6
d) 5
Discussion
Explanation: A number is divisible by 11 if the difference of the sum of the digits in the odd places and sum of the digits in even place is zero or divisible by 11.
Hence, (A + 8) - (3 + 1) = 0 or multiple of 11.
To get the difference 0 or multiple of 11, we need 7 at the place of A.
So, sum of odd place - sum of even place
= 15 - 4 = 11. And this is divisible by 11.
4. The greatest number which will divides: 4003, 4126 and 4249, leaving the same remainder in each case:
a) 43
b) 41
c) 45
d) None of these
Discussion
Explanation: Rule- Greatest number with which if we divide P, Q, R and it leaves same remainder in each case. Number is of form = HCF of (P - Q), (P - R)
Therefore, HCF of (4126 - 4003), (4249 - 4003) = HCF of 123, 246 = 41.
5. LCM of two numbers is 936. If their HCF is 4 and one of the numbers is 72, the other is
a) 62
b) 42
c) 52
d) None of these
Discussion
Explanation: Let two numbers be N1 and N2.
Now,
HCF of the numbers × LCM of the numbers = Multiplication of the numbers.
936 × 4 = 72 × N1
N1 = $$\frac{{936 \times 4}}{{72}}$$
N1 = 52
6. Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 12 noon, at what time will they beep again for first time?
a) 12:10 PM
b) 12:12 PM
c) 12:11 PM
d) 12:20 PM
Discussion
Explanation: They will ring together after,
LCM of 48 and 50 secs.
48 = 2 × 2 × 2 × 2 × 3;
50 = 2 × 5 × 5
LCM = 2 × 2 × 2 × 2 × 3 × 5 × 5 = 1200 secs
= 20 min.
They will beep together at 12:20
7. The sum of the digits of two-digit number is 10, while when the digits are reversed, the number decrease by 54. Find the changed number.
a) 28
b) 19
c) 37
d) 46
Discussion
Explanation: Let number be (10x + y)
According to question,
(10x + y) - (10y + x) = 54
10x - 10y + y - x = 54
9x - 9y = 54
x - y = 6 -------(i)
Sum of digits,
(x + y) = 10 ------- (ii)
(i) - (ii)
So, x - y - x - y = 6 - 10
-2y = -4
y = 2 and, x = 8
Then, the required number is
= (10y + x)
= 10 × 2 + 8
= 28
8. Find the HCF of (3125-1) and (335-1).
a) 35 - 1
b) 312 - 1
c) 34 - 1
d) None of these
Discussion
Explanation: Rule - The HCF of (am - 1) and (an - 1) is given by (aHCF of m, n - 1)
Thus for this question the answer is (35 - 1)
Since, 5 is the HCF of 35 and 125
9. The HCF of 2472, 1284 and a third number 'N' is 12. If their LCM is 23 × 32 × 5 × 103 × 107, then the number 'N' is:
a) 22 × 32 × 7
b) 22 × 33 × 10
c) 22 × 32 × 5
d) None of these
Discussion
Explanation: HCF of the numbers × LCM of the numbers = Multiplication of the numbers.
(12) × (23 × 32 × 5 × 103 × 107) = 2472 × 1284 × N
Hence,
N = $$\frac{{\{ \left( {12} \right) \times \left( {{2^3} \times {3^2} \times 5 \times 103 \times 107} \right)\} }}{{2472 \times 1284}}$$
N = 3 × 5
10. The greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32, 35 is.
a) 1120
b) 5600
c) 4714
d) 5200
Discussion
Explanation: LCM of 20, 28, 32, 35 will be the greatest number which is divisible by these numbers.
Firstly, we find LCM of 20, 28, 32, 35
20 = 2 × 2 × 5
28 = 2 × 2 × 7
32 = 2 × 2 × 2 × 2 × 2
35 = 5 × 7
LCM = 2 × 2 × 2 × 2 × 2 × 5 × 7 = 1120
Required greatest number which subtract from 5834 are divide by 20, 28, 32 and 35
= 5834 - 1120
= 4714
11. The sum of four consecutive two-digit odd numbers, when divided by 10, become a perfect square. Which of the following can possibly be one of these four numbers?
a) 21
b) 67
c) 25
d) 41
Discussion
Explanation: Using options,
We find that four consecutive odd numbers are 37, 39, 41 and 43
The sum of these 4 numbers is 160, when divided by 10 we get 16 which is a perfect square.
Thus, 41 is one of the odd numbers
12. If 381A is divisible by 9, find the value of the smallest natural number A?
a) 5
b) 8
c) 6
d) 9
Discussion
Explanation: A number is divisible by 9 when the sum of its digit is divisible by 9
So, (3 + 8 + 1 + A) = must be divisible by 9
Thus, smallest natural number be 6
(3 + 8 + 1 + 6) = 18, this is divisible by 9
13. The HCF of two numbers, each having three digits, is 17 and their LCM is 714. The sum of the numbers will be:
a) 289
b) 391
c) 221
d) 731
Discussion
Explanation: Let the numbers be 17x and 17y where x and y are co-prime.
LCM = 17xy
17xy = 714
xy = 42 = 6 × 7
→x = 6 and y = 7
x = 7 and y = 6
1st number = 17 × 6 = 102
2nd number = 17 × 7 = 119
Sum = 102 + 119 = 221
14. Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?
a) 1
b) 2
c) 4
d) 0
Discussion
Explanation: Any four digit number in which first two digits are equal and last two digits are also equal will be in the form 11 × (11a + b) i.e. it will be the multiple of 11 like 1122, 3366, 2244, . . . .
Now, let the required number be aabb.
Since aabb is a perfect square, the only pair of a and b that satisfy the above mentioned condition is a = 7 and b = 4
Hence, 7744 is a perfect square
15. A forester wants to plant 44 apples tree, 66 banana trees and 110 mango trees in equal rows (in terms of number of trees). Also, he wants to make distinct rows of tree (i.e. only one type of tree in one row). The number of rows (minimum) that required is:
a) 2
b) 3
c) 10
d) 11
Discussion
Explanation:
$$\eqalign{ & {\text{we}}\,{\text{first}}\,{\text{need}}\,{\text{to}}\,{\text{find}}\,{\text{the}}\,{\text{HCF}}\,{\text{of}}\,{\text{44,66,110}}. \cr & 44 = 2 \times 2 \times 11 \cr & 66 = 2 \times 3 \times 11 \cr & 110 = 2 \times 5 \times 11 \cr & {\text{HCF}} = 2 \times 11 = 22 \cr & {\text{Then,}}\,{\text{the}}\,{\text{required}}\,{\text{numbers}}\,{\text{of}}\,{\text{rows}}, \cr & = { {\frac{{44}}{{22}}} + {\frac{{66}}{{22}}} + {\frac{{110}}{{22}}} } \cr & = 10 \cr} $$
16. Which among $${2^{\frac{1}{2}}}$$, $${3^{\frac{1}{3}}}$$, $${4^{\frac{1}{4}}}$$, $${6^{\frac{1}{6}}}$$ and $${12^{\frac{1}{{12}}}}$$ is largest?
a) $${3^{\frac{1}{3}}}$$
b) $${4^{\frac{1}{4}}}$$
c) $${12^{\frac{1}{{12}}}}$$
d) $${2^{\frac{1}{2}}}$$
Discussion
Explanation: LCM in power 2, 3, 4, 6, 12 is 12
We multiplied the LCM to the power of the numbers.
$${2^{\frac{{1 \times 12}}{2}}},{3^{\frac{{1 \times 12}}{3}}},{4^{\frac{{1 \times 12}}{4}}},{6^{\frac{{1 \times 12}}{6}}}{\kern 1pt} {\text{and}}\,{12^{\frac{{1 \times 12}}{{12}}}}$$
We get,
= 26, 34, 43, 62, 12
= 64, 84, 64, 36, 12
Hence, greatest number would be $${3^{\frac{1}{3}}}$$
17. The last digit of the number obtained by multiplying the numbers 81 × 82 × 83 × 84 × 86 × 87 × 88 × 89 will be
a) 0
b) 6
c) 7
d) 2
Discussion
Explanation: The last digit of multiplication depends on the unit digit of (81 × 82 × 83 × 84 × 86 × 87 × 88 × 89) which is given by the remainder obtained on dividing it by 10.
$$\eqalign{ & \frac{{\left( {81 \times 82 \times 83 \times 84 \times 86 \times 87 \times 88 \times 89} \right)}}{{10}} \cr & {\text{We}}\,{\text{take}}\,{\text{individual}}\,{\text{remainder}}\,{\text{of}}\,{\text{each}}\,{\text{digit,}} \cr & \frac{{\left( {1 \times 2 \times 3 \times 4 \times 6 \times 7 \times 8 \times 9} \right)}}{{10}} \cr & {\text{Numbers}}\,{\text{multiplied}}, \cr & \frac{{\left( {24 \times 42 \times 72} \right)}}{{10}} \cr & {\text{Individual}}\,{\text{Remainder}}\,{\text{has}}\,{\text{been}}\,{\text{taken}}, \cr & \frac{{\left( {4 \times 2 \times 2} \right)}}{{10}} \cr & =\,\frac{{\left( {16} \right)}}{{10}} \cr & =\,6 \cr & {\text{Remainder}} = 6 \cr & {\text{So,}}\,{\text{the}}\,{\text{last}}\,{\text{digit}}\,{\text{will}}\,{\text{be}}\,6 \cr} $$
18. In a 4-digit number, the sum of the first two digits is equal to that of last two digits. The sum of the first and last digits is equal to third digit. Finally, the sum of the second and fourth digits is twice the sum of the other two digits. What is the third digit of the number?
a) 8
b) 1
c) 5
d) 4
Discussion
Explanation: Let the 1st, 2nd, 3rd and 4th digits be a, b, c and d respectively.
a + b = c + d ----------(i)
a + d = c ----------(ii )
b + d = 2(a + c) ----------(iii)
from eqn. (i) and (ii),
a + b = a + 2d
→b = 2d
and eqn (iii);
2d + d = 2(a + a + d)
→ 3d = 2(2a + d)
→ d = 4a
a = $$\frac{{\text{d}}}{4};$$
→ Now, from eqn. (ii),
a + d = $$\frac{{\text{d}}}{4}$$ + d = $$\frac{{{\text{5d}}}}{4}$$ = c
c = $$\frac{5}{{4{\text{d}}}}$$
The value of d can be either 4 or 8.
If d = 4, then c = 5
If d = 8, then c = 10
But the value of c should be less than 10
Hence, value of c would be 5
19. On a road three consecutive traffic lights change after 36, 42 and 72 seconds respectively. If the lights are first switched on at 9:00 AM sharp, at what time will they change simultaneously?
a) 9:08:04
b) 9:08:44
c) 9:08:24
d) None of these
Discussion
Explanation: LCM of 36, 42 and 72,
36 = 2 × 2 × 3 × 3
42 = 2 × 3 × 7
72 = 2 × 2 × 2 × 3 × 3
LCM = 2 ×2 × 2 × 3 × 3 × 7 = 504 seconds.
LCM of 36, 42 and 72 is 504
Hence, the lights will change simultaneously after 8 minutes and 24 seconds.
20. The rightmost non-zero digit of the number 302720 is
a) 1
b) 3
c) 7
d) 9
Discussion
Explanation: (30)2720, we can write it as[(30)4]680
= [(10 × 3)4]680
The right most non-zero digit depends on the unit digit of [(3)4]680
Unit digit of [(3)4]680,
= (81)680
The unit digit of 81 is 1 so any power of 81 will always give its unit digit as 1
Thus, required unit digit is 1
21. A two digit number ab is added to another number ba, which is obtained by reversing the digits then we get three digit number. Thus (a + b) equals to:
a) at least 18
b) 2ab
c) 2(a + b)
d) (a + b) ≥ 10
Discussion
Explanation: When two 2 digit numbers are added and the resultant value is a three digit number. It means there must be a carry over (i.e. The sum of unit digits be greater than 9. Similarly, the sum of the tens digit is also greater than 9)
Required numbers = 64 + 46 = 110
So, Option 'd' is correct.
22. A gardener plants his garden with 5550 trees and arranged them so that there is one plant more per row as there are rows then number of trees in a row is:
a) 56
b) 74
c) 76
d) 75
Discussion
Explanation: Let there be n rows, then number of trees in each row = (n + 1)
Total number of trees,
n × (n +1) = 5550
Now, at this moment this problem can be solved in two ways. First by finding the roots of quadratic equation. Second, by using the values from options.
74 × 75 = 5550
i.e. (n + 1) = 75
23. The sum of two numbers is 18. The greatest product of these two number can be:
a) 17
b) 80
c) 81
d) Can't determined
Discussion
Explanation: a + b = 18
Maximum of (a × b) will be only when a = b
Thus, a = b = 9
Maximum of (a × b) = 9 × 9 = 81.
24. The unit digit of (316)34n + 1 is :
a) 4
b) 5
c) 1
d) 7
Discussion
Explanation: The unit digit of (316)34n, depends on the power of 6.
See the pattern, 62 = 36
63 = 216
64 = 1296
Any power of 6 will give unit digit 6.
The unit digit of (316)34n always 6.
So, unit digit of (316)34n + 1 will be 7
25. A number when divided by 14 leaves reminder of 8, but when the same number is divided by 7, it will leave the remainder :
a) 3
b) 2
c) 1
d) 4
Discussion
Explanation: When the number is divided by 14 it gives a remainder of 8,
The number = 14N + 8 (14N is divisible by 14)
When same number is divided by 7 it will give remainder 1
26. The HCF and LCM of 24, 82, 162, 203 are :
a) 23, 32000
b) 24, 32000
c) 24, 25600
d) 22, 3200
Discussion
Explanation: HCF of 24, 82, 162, 203 = 24
LCM of 24, 82, 162, 203 = 24 = 28 × 125 = 32000
27. The four digit smallest positive number which when divided by 4, 5, 6 or 7, it leaves always the remainder as 3:
a) 1000
b) 1257
c) 1263
d) 1683
Discussion
Explanation: The least possible number = (LCM of 4, 5, 6 and 7) + 3
= 420 + 3
= 423
The next higher number is,
(420m + 3), now we put a least value of m such that
(420m + 3) ≥ 1000
At m = 3,
value = 420 × 3 + 3 = 1263
28. A string of length 221 metre is cut into two parts such that one part is $$\frac{9}{4}$$ th as long as the rest of the string, then the difference between the larger piece and the shorter piece is
a) 58 m
b) 53 m
c) 85 m
d) 76 m
Discussion
Explanation: Let one part of string is x.
x + $$\frac{{9{\text{x}}}}{4}$$ = 221 m
x = 68 meter
and, $$\frac{{9{\text{x}}}}{4}$$ = 153 m
Difference between two parts = 153 - 68 = 85 m
29. The sum of 100 terms of the series 1 - 3 + 5 - 7 + 9 - 11 .......... is:
a) 100
b) -200
c) 200
d) -100
Discussion
Explanation: 1 - 3 + 5 - 7 + 9 - 11 .......... + 197 - 199
= (- 2) + (-2) + (-2) + .......... + (-2) (50 times)
= 50 × (-2) = -100
30. The remainder of $$\frac{{{6^{36}}}}{{215}}:$$
a) 0
b) 1
c) 2
d) 3
Discussion
Explanation:
$$\eqalign{ & \frac{{ {{6^{36}}} }}{{215}},\,{\text{can}}\,{\text{be}}\,{\text{written}}\,{\text{as}} \cr & \frac{{{{\left( {{6^3}} \right)}^{12}}}}{{215}} \cr & \frac{{{{216}^{12}}}}{{215}},\,[216\,{\text{on}}\,{\text{divided}}\,{\text{by}}\,215,\,{\text{gives}}\,{\text{remainder}}\,1] \cr & \frac{{{1^{12}}}}{{215}} \cr & {\text{The}}\,{\text{remainder}}\,{\text{will}}\,{\text{be}}\,1 \cr} $$
31. Three-fifth of a number is equal to 70% of another number. What is the ratio between the first number and second number ?
a) 4 : 7
b) 3 : 4
c) 7 : 8
d) 7 : 6
Discussion
Explanation:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{two}}\,{\text{numbers}}\,{\text{be}}\,X\,{\text{and}}\,Y \cr & X \times {\frac{3}{5}} = 70\% \,of\,Y \cr & \frac{{3X}}{5} = \frac{{70Y}}{{100}} \cr & \frac{X}{Y} = \frac{{ {70 \times 5} }}{{ {100 \times 3} }} = \frac{7}{6} \cr & X:Y = 7:6 \cr} $$
32. A virus known to cause a deadly disease, is also a rapid multiplier. It is known that it can double itself every 30 minutes. If a container is completely full of this virus after 50 hours, when was the container half empty?
a) 35 hours
b) 49.5 hours
c) 46 hours
d) 20 hours
Discussion
Explanation: In every 30 minutes, Virus doubles it self. It means in 49.5 hours, it would be half empty and it will be full in 50 hours.
33. The LCM of two number is 45 times their HCF. If one of the numbers is 125 and the sum of HCF and LCM is 1150, the other number is:
a) 215
b) 220
c) 225
d) 235
Discussion
Explanation: LCM = 45 HCF
LCM + HCF = 1150
45 × HCF + HCF = 1150
46 × HCF = 1150
HCF = 25
LCM = 45 × 25 = 1125
Now, use the formula,
1st number × second Number = LCM × HCF
125 × second number = 1125 × 25
Second Number = 225
34. The smallest number of four digits which on division by 4, 6, 10 and 15 leaves a remainder 2 in each case is:
a) 1020
b) 1022
c) 1024
d) 1040
Discussion
Explanation: First of all,we find the LCM of 4, 6, 10 and 15
LCM of 4, 6, 10, 10 = 60
The smallest four digit no. is 1000. We divide it by 60
$$\frac{{1000}}{{60}}$$ It leaves remainder 40
Now, the smallest four digit no which is divisible by 4, 6, 10, 15 is,
1000 + (60 - 40) = 1020
Required number (as it gives remainder 2 always) would be = 1020 + 2 = 1022
35. One-third of a tower is painted black, $$\frac{5}{{11}}$$th of the remaining part is painted red and the rest is painted white. If the white part measures 60 ft.,the total height of the tower is-
a) 3100 ft.
b) 154 ft.
c) 165 ft.
d) 110 ft.
Discussion
Explanation: Let the tower height will be x ft.
Black part will be = $$\frac{1}{3}x$$
Remaining part = $$x - \frac{1}{3}x = \frac{2}{3}x$$
Red part will be = $$\frac{2}{3}x \times \frac{5}{{11}} = \frac{{10}}{{33}}x$$
White part will be = $$\frac{2}{3}x - \frac{{10}}{{33}}x = \frac{{12x}}{{33}}$$
$$\frac{{12x}}{{33}}$$ = 60
x = 165 ft.
36. The greatest number will divide 3026 and 5053 leaving remainders 11 and 13 respectively?
a) 15
b) 30
c) 40
d) 60
Discussion
Explanation: Required Number can be given by:
HCF of (3026 - 11) and (5053 - 13)
HCF of 3015 and 5040 = 15
To find HCF, we break the given numbers in their prime Factors
3015 = 3 × 3 × 5 × 67
5040 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7
We take common multiples in these two given numbers to get required HCF
And Common multiples are: 3 × 5
Required HCF = 15
37. The difference between three times and seven times of a number comes to 36. What is the number?
a) 6
b) 8
c) 9
d) 10
Discussion
Explanation: Let number be X
7X - 3X = 36
4X = 36
X = 9
Required Number is 9
38. Amitabh picks a random integer between 1 and 999, doubles it and gives the result to Sashi. Each time Sashi gets a number from Amitabh, he adds 50 to the number, and gives the result back to Amitabh, who doubles the number again. The first person, whose result is more than 1000, loses the game. Let 'X' be the smallest initial number that results in a win for Amitabh. The sum of the digit of 'X' is:
a) 3
b) 5
c) 7
d) 9
Discussion
Explanation: Assume x is the required number.
Step 1 Amitabh 2x Sashi 2x + 50
Step 2 Amitabh 4x + 100. Sashi 4x + 150
Step 3 Amitabh 8x + 300. Sashi 8x + 350
Step 4 Amitabh 16x + 700 Sashi 16x + 750
16x + 750 > 1000
And, x can be 16
7 is the sum
39. Prof. Suman takes a number of quizzes for a course. All the quizzes are out of 100. A student can get an A grade in the course if the average of her scores is more than or equal to 90. Grade B is awarded to a student if the average of her scores is between 87 and 89 (both included). If the average is below 87, the student gets a C grade. Ramesh is preparing for his last quiz and he realizes that he must score a minimum of 97 to get an A grade. After the quiz, he realizes that he will score 70, and he will just manage a B. how many quizzes did Prof. Suman take?
a) 6
b) 7
c) 8
d) 9
Discussion
Explanation: The difference between just getting an A grade and just getting a B grade would be equal to (number of quizzes × 3).
The difference between the required score to manage an A grade and the score achieved to just manage a B grade as given in the problem’s information is equal to 97 – 70 = 27
number of quizzes × 3 = 27
Number of quizzes is 9
40. A student got twice as many sums wrong as he got right. If he attempted 48 sums in all, how many did he solve correctly?
a) 12
b) 16
c) 24
d) 18
Discussion
Explanation: Let he has solved correctly X no. of sums. Therefore incorrect no. of sums = 2X
X + 2X = 48
3X = 48
X = 16 sums he has done correctly.
41. The number of prime factors in the expressions 64 × 86 × 108 × 1210 is:
a) 80
b) 64
c) 72
d) 48
Discussion
Explanation: 64 × 86 × 108 × 1210
= (2 × 3)4 × (23 )6 × (2 × 5)8 × (22 × 3 )10
= 24 × 34 × 218 × 28 × 58 × 220 × 310
= 250 × 314 × 58
The total prime factors,
= 50 + 14 + 8 [By adding maximum power of prime factors.]
= 72
42. x is five digit number. The digit in ten thousands place is 1. the number formed by its digits in units and ten places is divisible by 4. The sum of all the digits is divisible by 3. If 5 and 7 also divide x, then x will be.
a) 14020
b) 12060
c) 10020
d) 10080
Discussion
Explanation: Let the digits of x be
x = abcde
x = 1bcde [Given ten thousands place is 1.]
Now we can check the options as given that the sum of the all digit is divisible by 3.
10080 is the only number given in the option which satisfies all the given conditions.
43. A man sells chocolates which are in the boxes. Only either full box or half a box of chocolates can be purchased from him. A customer comes and buys half the number of boxes which the seller had plus half box more. A second customer comes and purchases half the remaining number of boxes plus half a box. After this the seller is left with no chocolate boxes. How many chocolate boxes the seller had initially?
a) 2
b) 3
c) 4
d) 3.5
Discussion
Explanation: The best way to go through the options
Let there are initially 3 boxes then,
1st customer gets = $$\frac{3}{2}$$ + $$\frac{1}{2}$$ = 2
Remaining boxes = 3 - 2 = 1
2nd customer = $$\frac{1}{2}$$ + $$\frac{1}{2}$$ = 1
Option 'b' is correct.
44. If x + y + z = 0, then x3 + y3 + z3 is equal to :
a) 0
b) 3xyz
c) $$\frac{{{\text{xy}} + {\text{yz}} + {\text{zx}}}}{{{\text{xyz}}}}$$
d) xyz(xy + yz + zx)
Discussion
Explanation:
x + y + z = 0
Cubing both side,
(x + y + z)3 = 0
x3 + y3 + z3 - 3xyz = 0 [using formula]
x3 + y3 + z3 = 3xyz
45. To write all the page numbers of a book, exactly 136 times digit 1 has been used. Find the number of pages in the book.
a) 190
b) 195
c) 210
d) 220
Discussion
Explanation: From 1- 99 digit 1 is used 20 times. And From 100 - 199, 1 is used 120 times
So, from 1 to 199, 1 is used,
20 + 120 = 140 times
We need 136. So leave 199, 198, 197 and 196
Required pages = 195
46. Given, N = 98765432109876543210 ..... up to 1000 digits, find the smallest natural number n such that N + n is divisible by 11.
a) 2
b) 3
c) 4
d) 5
Discussion
Explanation: For a no. to be divisible by 11,
Sum(odd digit nos) - Sum(even digit nos) = 0 or divisible by 11
If we look at 9876543210, the difference we get is 5
i.e. [(9 + 7 + 5 + 3 + 1) - (8 + 6 + 4 + 2 + 0) = 5]
The series is up to 1000 digit,
That means, $$\frac{{1000}}{{10}}$$ = 100 time 5,
then the difference will be 5 × 100 = 500
In order for the difference to be divisible by 11, we need to add 5 and the no will become 505
505 is divisible by 11
47. The remainder when 40 + 41 + 42 + 43 + ........ + 440 is divided by 17 is:
a) 0
b) 16
c) 4
d) None of these
Discussion
Explanation: Let S be the sum of the expression,
S = 40 + 41 + 42 + 43 + ........ + 440
S = (1 + 4 + 16 + 64) + 44 (1 = 4 + 16 + 64) + ...... + 436 + 440
Since, (1 + 4 + 16 + 64) = 85, is divisible by 17. Hence, except 440 remaining expression is divisible by 17.
$$\frac{{{4^{40}}}}{{17}} \to \frac{{{{\left( {{4^4}} \right)}^{10}}}}{{17}} \to \frac{{{1^{10}}}}{{17}}$$
Remainder = 1
48. The remainder when 30 + 31 + 32 + 33 + . . . . . . . + 3200 is divided by 13 is:
a) 0
b) 4
c) 3
d) 12
Discussion
Explanation:
$$\eqalign{ & {\text{The}}\,{\text{given}}\,{\text{expression}}\,{\text{is}}\,{\text{in}}\,{\text{GP}}\,{\text{series}} \cr & S = {3^0} + {3^1} + {3^2} + {3^3} + ........ + {3^{200}} \cr & S = {\frac{{ {{3^0} \times \left( {{3^{201}} - 1} \right)} }}{{ {3 - 1} }}} \cr & S = \frac{{ {{3^{201}} - 1} }}{2} \cr & S = \frac{{ {{{\left( {{3^3}} \right)}^{67}} - {1^3}} }}{2} \cr & S = \frac{{ {{{27}^{67}} - {1^3}} }}{2} \cr} $$
Since, (An - Bn) is divisible by (A - B), So, (2767 - 13) is divisible by (27 - 1) = 26
Hence, Expression is also divisible by 13 as it is divisible by 26
Given expression is divisible by 13 so the remainder will be 0
49. The distance between the house of Rajan and Raman is 900 km and the house of former is at 100th milestone where as the house of Raman is at 1000th milestone. There are total 901 milestone at a regular interval of 1 km each. When you go to Raman's house from the house of Rajan which are on same highway, you will find that if the last digit (i.e. unit digit) of 3 digit number on every milestone is same as the first (i.e. hundreds digit) of the number on the next milestone is same, then these milestones must be red colour and rest will be of black. Total number of red colour milestone is:
a) 179
b) 90
c) 189
d) 100
Discussion
Explanation: Rajan (100)____________________ (1000) Raman
These numbers are:
(101, 102), (111, 102), (121, 122), (131,132).........(202, 203), (212, 213), ....... (303, 304), (313, 314) ......... (808, 809)......... (989, 990)
There are total 20 × 7 + 21 + 18 = 179, mile stones which are red, because the hundreds digit on the next milestone is same as the unit digit of previous milestone.
50. Half way through the journey from Delhi to Lahore Atalji began to look out of the window of the Samjhauta Express and continued it until the distance which was remained to cover was half of what he has covered. Now at this time how much distance he has to cover?
a) $$\frac{2}{5}$$
b) $$\frac{1}{4}$$
c) $$\frac{1}{3}$$
d) $$\frac{1}{6}$$
Discussion
Explanation: Since, he has covered twice the distance which yet he has to cover. It means he has covered $$\frac{2}{3}$$ of the whole journey and remaining journey is $$\frac{1}{3}$$ rd.
51. Ram left $$\frac{1}{3}$$ of his property to his widow and $$\frac{3}{5}$$ of the remainder to his daughter. He gave the rest to his son who received Rs. 6,400. How much was his original property worth?
a) Rs. 16,000
b) Rs. 24,000
c) Rs. 30,000
d) Rs. 32,000
Discussion
Explanation:
$$\eqalign{ & {\text{Let}}\,{\text{total}}\,{\text{property}}\,{\text{of}}\,{\text{Ram}} = x \cr & {\text{His}}\,{\text{widow}}\,{\text{gets}} \cr & \frac{1}{3}\,of\,x = \frac{x}{3} \cr & {\text{His}}\,{\text{daughter}}\,{\text{gets}}\, \cr & = \frac{3}{5}\,of\,\left[ {x - {\frac{x}{3}} } \right] \cr & = \frac{3}{5} \times {\frac{{2x}}{3}} \cr & = \frac{{2x}}{5} \cr & {\text{His}}\,{\text{son}}\,{\text{gets}}\,{\text{his}}\,{\text{rest}}\,{\text{property}} \cr & {\text{His}}\,{\text{son's}}\,{\text{property}} \cr & = x - \left[ { {\frac{x}{3}} + {\frac{{2x}}{5}} } \right] \cr & 6400 = x - {\frac{{11x}}{{15}}} \cr & 6400 = \frac{{ {15x - 11x} }}{{15}} \cr & 4x = 6400 \times 15 \cr & x = 1600 \times 15 \cr & x = 24000 \cr & {\text{Total}}\,{\text{Property}} = 24000 \cr} $$
52. If x : y be the ratio of two whole numbers and z be their HCF, then the LCM of those two numbers is:
a) $${\text{yz}}$$
b) $$\frac{{{\text{xz}}}}{{\text{y}}}$$
c) $$\frac{{{\text{xy}}}}{{\text{z}}}$$
d) $${\text{xyz}}$$
Discussion
Explanation: Ratio of the numbers = x : y
HCF of the numbers = z
So, z is the common factor of the numbers
Then, First number = xz
Second Number = yz
First Number × Second Number = HCF and LCM of the numbers
xzyz = z × LCM
LCM = xyz
53. The sum of the digits of a two-digits numbers is 12 and when the digits of the to digit number are interchanged, the new number is 36 more than the original. What is the Original number?
a) 93
b) 48
c) 39
d) 84
Discussion
Explanation: Let the Original number is (X + 10Y)
Sum of the digit of the number = 12
X + Y = 12 ----------- (1) When digit interchanged, number become 36 more than original number,
10X + Y = 10Y + X + 36
X - Y = 4 ------------------- (2)
On solving equation (1) and (2),
X = 8
Y = 4
Original number,
= X + 10Y = 8 + 10 × 4 = 48
54. A certain number of the capsule were purchased for Rs. 176. Six more capsule could have been purchased in the same amount if each capsule were cheaper by Rs. 3. What was the number of capsules purchased?
a) 13
b) 16
c) 17
d) 8
Discussion
Explanation: Let Price of each capsule were Rs. X, originally.
So, Number of capsule purchased originally,
= $$ \frac{{176}}{{\text{X}}}$$
If price of capsule is Rs. 3 less then 6 more capsule were purchased. In this case, the number of capsules,
= $$ \frac{{176}}{{{\text{X}} - 3}}$$
$$\eqalign{ & \frac{{176}}{{{\text{X}} - 3}} - \frac{{176}}{{\text{X}}} = 6 \cr & \frac{{176{\text{X}} - 176{\text{X}} + 528}}{{{\text{X}} \times \left( {{\text{X}} - 3} \right)}} = 6 \cr & 6{{\text{X}}^2} - 18{\text{X}} - 528 = 0 \cr & {{\text{X}}^2} - 3{\text{X}} - 88 = 0 \cr & {{\text{X}}^2} - 11{\text{X}} + 8{\text{X}} - 88 = 0 \cr & \left( {{\text{X}} - 11} \right)\left( {{\text{X}} + 8} \right) = 0 \cr} $$
Either, X = 11 Or, X = -8 (Price cannot be negative)
Thus, X = Rs. 11
Number of the capsule purchased = $$\frac{{176}}{{11}}$$ = 16
55. An even number can be expressed as the square of an integer as well as cube of another integer. Then the number has to be necessarily divisible by?
a) 32
b) 64
c) 128
d) Both (a) and (b)
Discussion
Explanation: Check the option. 32 and 128 are neither of a square and nor cube root. 64 is the square of 8 and a cube of 4
56. The remainder when (1213 + 2313) is divided by 11.
a) 0
b) 1
c) 2
d) 3
Discussion
Explanation:
$$\eqalign{ & \frac{{{{12}^{13}}}}{{11}}\,{\text{gives}}\,{\text{Remainder}}\,1 \cr & {\text{It}}\,{\text{can}}\,{\text{be}}\,{\text{written}}\,{\text{as}}\, \cr & \frac{{\left( {12 \times 12 \times 12 \times 12\,........\,13\,\text{times}} \right)}}{{11}} \cr & {\text{On}}\,{\text{dividing}}\,{\text{it}}\,{\text{gives}}\,{\text{remainder}}\,{\text{1,}}\,{\text{each}}\,{\text{time}}. \cr & 1 \times 1 \times 1 \times 1\,.........\,13\,{\text{times}} \cr & {\text{So,}}\,{\text{final}}\,{\text{remainder}}\,{\text{will}}\,{\text{be}}\,1 \cr & \frac{{{{23}^{13}}}}{{11}} \Rightarrow \,{\text{Remainder}}\,1 \cr & {\text{The}}\,{\text{remainder}}\,{\text{of}}\,\frac{{\left( {{{12}^{13}} + {{23}^{13}}} \right)}}{{11}} \cr & = \left( {1 + 1} \right) \cr & = 2 \cr} $$
57. The remainder when 757575 is divide by 37.
a) 0
b) 1
c) 5
d) 7
Discussion
Explanation: $$\frac{{{{75}^{{{75}^{75}}}}}}{{37}}$$
When 75 is divided by 37, it leaves remainder of 1.
The expression will become,
$$\frac{{{1^{{{75}^{75}}}}}}{{37}}$$
Now, for any power of 1, we always get 1 and expression becomes $$\frac{1}{{37}}$$ that leaves remainder 1
58. The product of the digits of a three digit number is a perfect square and perfect cube both is :
a) 126
b) 256
c) 18
d) None of these
Discussion
Explanation: The only possible number which is both perfect square and perfect cube is 729 which is cube of 9 and square of 27
7 × 2 × 9 = 126
59. The LCM of two numbers is 1020 and their HCF is 34 the possible pair of number is:
a) 255, 34
b) 102, 204
c) 204, 170
d) None of these
Discussion
Explanation: Check options one by one: LCM of 204 and 170 = 1020.
60. Sunny gets $$\frac{7}{9}$$ times as many marks in QA as in ENGLISH. If his total combined marks in both the papers is 90. His marks in QA is:
a) 50
b) 60
c) 70
d) 80
Discussion
Explanation:
$$\eqalign{ & {\text{Total}}\,{\text{Marks}} = 90 \cr & {\text{Marks}}\,{\text{in}}\,{\text{QA}} = \frac{7}{9}\,\text{of}\,\,90 \cr & = \frac{{\left( {7 \times 90} \right)}}{9} \cr & = 7 \times 10 \cr & = 70 \cr} $$
61. In a certain series, each number except the first and second is obtained by adding the previous two numbers. If the first no is 2 and sixth no is 26, then the seventh number is:
a) 52
b) 46
c) 40
d) 42
Discussion
Explanation: 1st no. = 2
2nd no. = x (let)
3rd no. = x + 2
4th no. = x + x + 2 = 2x + 2
5th no. = x + 2 + 2x + 2 = 3x + 4
6th no.,
3x + 4 + 2x + 2 = 26
5x = 20
x = 4
7th no. = 26 + 3x + 4 = 26 + 3 × 4 + 4 = 42
62. If (-1)n + (-1)4n = 0, then n is
a) Any positive integer
b) Any negative integer
c) Any odd natural number
d) Any even natural number
Discussion
Explanation: Putting n = 1, 2, 3, 4, 5 . . . .
n = 1 in the given equation
(-1)1 + (-1)4 × 1 = 0
-1 + 1 = 0
n = 2,
(-1)2 + (-1)4 × 2 = 0
1 + 1 ≠ 0
n = 3,
-1 + 1 = 0
We will see that the given condition will be satisfy only for any odd natural number.
63. Sunil entered a shopping center and spent one half of the money that he had. When he finished his purchase he found that he had as many paise as he had rupees and half as many rupees as he had paise when he went. How much money did he have when he entered?
a) Rs. 75.50
b) Rs. 98.98
c) Rs. 99.98
d) None of these
Discussion
Explanation: Here we have to verify the given two conditions for the given options.
Condition :1
As given, after finished his purchase he had the rupees as much as the paise when he went.
Condition :2
After finishing purchase, he had Half as many rupees as he had paise when he went.
Solve it through option checking method.
Case :1
Now suppose the answer is option c
Then he spent Rs. $$\frac{{99.98}}{2}$$ = Rs. 49.99
Condition :1
Here we have 99 paise so the rupees should be 99 when he went. He initially had 99 Rs. Hence condition 1 is satisfied.
Condition :2
He had 98 paise to begin with. So the rupees he spent should be $$\frac{{98}}{2}$$ = Rs. 49 as per condition 2.
It turns out that the rupees actually he spent was 49.
Therefore, condition 2 is satisfied as well.
64. 11 * X = $$\frac{{693}}{9}$$
a) 3
b) 5
c) 7
d) 9
Discussion
Explanation:
$$\eqalign{ & X = \frac{{ {693} }}{{ {9 \times 11} }} \cr & X = 7 \cr} $$
65. What is the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?
a) 840
b) 841
c) 820
d) 814
Discussion
Explanation: Length of the room
= 15 m 17 cm
= 15 × 100 + 17
= 1517 cm.
Breadth of the room
= 9 m 2 cm
= 902 cm.
The HCF of the 1517 and 902 will be size of square tiles.
HCF of 1517 and 902 = 41 cm.
Area of the room
= length × breadth
= 1517 × 902 cm2
Area of tiles = 41 × 41 cm2
Number of tiles required
= $$\frac{{1517 \times 902}}{{41 \times 41}}$$
= 814 tiles
66. Three numbers which are co-prime to each other are such that the product of the first two is 119 and that of the last two is 391. What is the sum of the three numbers?
a) 47
b) 43
c) 53
d) 51
Discussion
Explanation: Since the numbers are co-prime, their HCF = 1
Product of first two numbers = 119
Product of last two numbers = 391
The middle number is common in both of these products.
Hence if we take HCF of 119 and 391, we get the common middle number.
HCF of 119 and 391 = 17
So, Middle Number = 17
First Number = $$\frac{{119}}{{17}}$$ = 7
Last Number = $$\frac{{391}}{{17}}$$ = 23
Sum of the three numbers
= 7 + 17 + 23
= 47
67. A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they again at the starting point ?
a) 36 minutes 32 seconds
b) 36 minutes 22 seconds
c) 36 minutes 12 seconds
d) 46 minutes 12 seconds
Discussion
Explanation: LCM of 252, 308 and 198 = 2772
They all will be again at the starting point after 2772 seconds or 46 minutes 12 seconds.
68. What is the least number when divided by 8, 12, 20 and 30 leaves a remainder 4 and when divided by 21 leaves a remainder 7 ?
a) 124
b) 360
c) 364
d) Cannot be determined
Discussion
Explanation: Lcm of 8, 12,20,30 is 120. As they leave a remainder 4
So it should be 120 + 4. But when being divided by 21, remainder is 7
So the number will be exactly divisible by (21 + 7) = 28.
The number will be in the form of (120k + 4) which is divisible by 28
Where k is any positive integer. Now, if we divide 120 by 28, the remainder is 8
We need a remainder of 24, because (24 + 4) = 28
Now (8 × 3) = 24
k = 3
The number would be (120 × 3) + 4 = 364
69. Six bells start ringing together and ring at intervals of 4, 8, 10, 12, 15 and 20 seconds respectively. how many times will they ring together in 60 minutes ?
a) 31
b) 15
c) 16
d) 30
Discussion
Explanation: LCM of 4, 8, 10, 12, 15 and 20 = 120
120 seconds = 2 minutes
All the six bells will ring together in every 2 minutes
Number of times they will ring together in 60 minutes
= $$1 + \frac{{60}}{2}$$
= 31
70. What is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder?
a) 1108
b) 1683
c) 2007
d) 3363
Discussion
Explanation: LCM of 5, 6, 7 and 8 = 840
Hence the number can be written in the form (840k + 3) which is divisible by 9
If k = 1, number = (840 × 1) + 3 = 843 which is not divisible by 9
If k = 2, number = (840 × 2) + 3 = 1683 which is divisible by 9
1683 is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder
71. The remainder when 1010 + 10100 + 101000 + . . . . . . + 101000000000 is divided by 7 is
a) 0
b) 1
c) 2
d) 5
Discussion
Explanation: Number of terms in the series = 10.
(We can get it easily by pointing the number of zeros in power of terms.
In 1st term number of zero is 1, 2nd term 2, and 3rd term 3 and so on)
$$\frac{{{{10}^{10}}}}{7},$$ Written as, $$\frac{{{{\left( {7 + 3} \right)}^{\left( {4 \times 2 + 2} \right)}}}}{7}$$
The remainder will depend on $$\frac{{{3^2}}}{7}$$
So, remainder will be 2
$$\eqalign{ & \frac{{{{10}^{1000}}}}{7},\,{\text{remainder}} = 2 \cr & \frac{{{{10}^{10000}}}}{7},\,{\text{remainder}} = 1 \cr} $$
We get alternate 2 and 1 as remainder, five times each.
Required remainder is given by
$$\eqalign{ & \frac{{\left( {2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1} \right)}}{7} \cr & = \frac{{15}}{7} \cr} $$
Remainder when 15 is divided by 7 = 1
72. There are 154 beads in a rosary and all are coloured, either RED or BLUE or GREEN. The number of blue ones is three less than red and five more than green. The number of Red beads are:
a) 55
b) 47
c) 45
d) 52
Discussion
Explanation: Let the number of red beads = x
Blue beads = (x - 3)
Green beads = (x - 8)
x + x - 3 + x - 8 = 154
3x - 11 = 154
x = 55
73. Recently, a small village, in Tamilnadu where only male shepherd reside with four sheep each, was devastated by Tsunami waves. Therefore 8 persons and 47 sheep were found to be dead and the person who luckily survived, left the village with one sheep each since 21 sheep were too injured to move so have left on their on luck, in the village. The number of sheep which were earlier in the village was :
a) 84 sheep
b) 120 sheep
c) 80 sheep
d) 90 sheep
Discussion
Explanation: Let there be x people in the village so, then 4x sheep must be there.
No, number of person leaving the village,
= (x - 8) [as 8 person died in Tsunami]
And the number of sheep to be carried with the survived people,
4x - (47 + 21) = x - 8. [Here we are equating no. of alive sheep and no. of alive person. As 1 person has 4 sheep means 1 person = 4 sheep.]
3x = 60
x = 20
4x = 80
So, 80 sheep
74. A boy appeared in CAT for four consecutive year years, but coincidentally each time his net score was 75. He told me that there was $$\frac{1}{3}$$rd negative marking for every wrong answer and 1 mark was allotted for every correct answer. He has attempted all the questions every year, but certainly some answers have been wrong due conceptual problem. Which is not the total number of questions asked for CAT in any year, in that period?
a) 231
b) 163
c) 150
d) 123
Discussion
Explanation: Since 75 is an integer, he must have to do (75 + 4x) problems, where x is a whole number. Now we put the value of x (0, 1, 2, 3, 4, 5 .........) to get the correct answer.
As he gets the net marks in integer it means when he does 3n question wrong, he losses actually 4n marks.
75. The highest power of 17 which can divide exactly the following:
(182 - 1) (184 - 1) (186 - 1) (188 - 1) . . . . (1816 - 1) (1818 - 1) is :
a) 1
b) 17
c) 9
d) 13
Discussion
Explanation: Total number of terms in expression = 9
(182 - 1) = (18 - 1) × (18 + 1)
(184 - 1) = (182 + 1) × (182 - 1) = (182 + 1) × (18 - 1) × (18 + 1)
(186 - 1) = [(183)2 -1] = (183 + 1) (183 - 1) = 17 × k and so on.
There will be 9 times 17 in the whole expression as each term of expression gives one 17. Therefore for maximum power of 17 will be 9.
76. The owner of a local jewellery store hired 3 watchmen to guard his diamonds, but a thief still got in and stole some diamonds. On the way out, the thief met each watchman, one at a time. To each he gave $$\frac{1}{2}$$ of the diamonds he had then and 2 more besides. He escaped with one diamond. How many did he steal originally?
a) 36
b) 25
c) 640
d) None of these
Discussion
Explanation: At last thief is left with one diamond.
Hence, the number of diamonds before he gave some diamonds to the third watchman,
$$\eqalign{ & x - \left( { {\frac{x}{2}} + 2} \right) = 1 \cr & {\kern 1pt} \frac{{ {x - 4} }}{2} = 1 \cr & {\kern 1pt} x = 6 \cr} $$
Hence, he had 6 diamonds before he gave 5 to the third watchman.
Similarly number of diamonds before giving to second watchman,
$$\eqalign{ & \frac{{ {x - 4} }}{2} = 6 \cr & {\kern 1pt} x = 16 \cr} $$
And number of diamonds before giving to the first watchman,
$$\eqalign{ & \frac{{ {x - 4} }}{2} = 16 \cr & {\kern 1pt} x = 36 \cr} $$
The thief has stolen 36 diamonds originally
77. Which one of the following is not a prime number?
a) 31
b) 61
c) 71
d) 91
Discussion
Explanation: 91 is divisible by 7. So, it is not a prime number.
78. The remainder , when (22225555 + 55552222) is divided by 7, is
a) 4
b) 5
c) 0
d) 2
Discussion
Explanation: an + bn is always divisible by (a + b) when n is odd.
(22225555 + 55552222) is always divisible by (2222 + 5555) = 7777
And 7777 is multiple of 7,
(22225555 + 55552222) is divisible by 7
79. A young girl counted in the following way on the fingers of her left hand. She started calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8, thumb 9 and then back to the index finger for 10, middle finger for 11 and so on. She counted up 1994. She ended on her
a) Thumb
b) Index finger
c) Middle finger
d) Ring finger
Discussion
Explanation: If the girl counts the way as given in the question,
The counting serial for thumb will be 1, 9, 17, 25 . . . . . .
Number 1992 ( as 1992 is divisible by 8, common difference of series formed by counting) will also fall on thumb.
Hence, Number 1994 will end on her middle finger.
80. How many pairs of natural numbers is there the difference of whose squares are 45.
a) 1
b) 2
c) 3
d) 4
Discussion
Explanation: (x2 - y2) = 45
(x - y)(x + y) = 45
The factors of 45 possibles are 15, 3, 9, 5, 1 and 45
Hence, numbers are 9 and 6 or 7 and 2 or 23 and 22
Number of such pairs = 3
81. If 146 Is divisible by 5n, and then find the maximum value of n.
a) 34
b) 35
c) 36
d) 37
Discussion
Explanation:
$$\eqalign{ & = {\frac{{146}}{5}} + {\frac{{146}}{{{5^2}}}} + {\frac{{146}}{{{5^3}}}} \cr & = 29 + 5 + 1 \cr & = 35 \cr} $$
Note: We have taken integral value only, not the fractional. For example $$\frac{{146}}{5}$$ = 29.2 but we have taken 29 and so on.
82. Three friends, returning from a movie, stopped to eat at a restaurant. After dinner, they paid their bill and noticed a bowl of mints at the front counter. Sita took $$\frac{1}{3}$$ of the mints, but returned four because she had a monetary pang of guilt. Fatima then took $$\frac{1}{4}$$ of what was left but returned three for similar reasons. Eswari then took half of the remainder but threw two back into the bowl. The bowl had only 17 mints left when the raid was over. How many mints were originally in the bowl?
a) 38
b) 31
c) 41
d) None of these
Discussion
Explanation:
$$\eqalign{ & {\text{Number of mint before Eswari has taken}}, \cr & = \left( {x - {\frac{x}{2}} } \right) + 2 = 17 \cr & x = 30 \cr & {\text{Number of mint before Fatima has taken}}, \cr & = \left( {x - {\frac{x}{4}} } \right) + 3 = 30 \cr & x = 36 \cr & {\text{Number of mint before Sita has taken}}, \cr & = \left( {x - {\frac{x}{3}} } \right) + 4 = 36 \cr & x = 48 \cr & {\text{There}}\,{\text{were}}\,{\text{48}}\,{\text{mints}}\,{\text{originally}}{\text{.}} \cr} $$
83. Some birds settled on the branches of a tree. First, they sat one to a branch and there was one bird too many. Next they sat two to a branch and there was one branch too many. How many branches were there?
a) 3
b) 4
c) 5
d) 6
Discussion
Explanation: When the birds sat one on a branch, there was one extra bird.
When they sat Two to a branch one branch was extra.
Go through the options one by one;
Checking option (a);
If there were 3 branches, there would be 4 birds.
(This would leave one bird without branch as per the question)
When 4 birds would sit Two to a branch there would be one branch free
(as per the question).
The option 'a' is correct.
84. If we divide the unknown two-digit number by the number consisting of the same digits written in the reverse order, we get 4 as quotient and 3 as remainder. If we divide the required number by sum of its digits, we get 8 as a quotient and 7 as a remainder. Find the number?
a) 81
b) 91
c) 71
d) 72
Discussion
Explanation: Go through the options one by one,
Checking option (c);
$$\frac{{71}}{{17}}$$ = 4 quotient and remainder 3.
$$\frac{{71}}{8}$$ = 8 quotient and remainder 7 as remainder.
85. The last three-digits of the multiplication 12345 × 54321 will be
a) 865
b) 745
c) 845
d) 945
Discussion
Explanation: If we multiply the last three digits of each terms we get the last three digits.
345 × 321 = 110745
The last three digits are 745
86. Find the least number which will leaves remainder 5 when divided by 8, 12, 16 and 20
a) 240
b) 245
c) 265
d) 235
Discussion
Explanation: We have to find the Least number, therefore we find out the LCM of 8, 12, 16 and 20.
8 = 2 × 2 × 2;
12 = 2 × 2 × 3;
16 = 2 × 2 × 2 × 2;
20 = 2 × 2 × 5;
LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240;
This is the least number which is exactly divisible by 8, 12, 16 and 20
Required number which leaves remainder 5 is,
240 + 5 = 245
87. 76n- 66n, where n is an integer >0, is divisible by
a) 13
b) 127
c) 559
d) All of these
Discussion
Explanation:
$$\eqalign{ & {7^{6n}} - {6^{6n}} \cr & = {7^6} - {6^6} \cr & = {\left( {{7^3}} \right)^2} - {\left( {{6^3}} \right)^2} \cr & = \left( {{7^3} - {6^3}} \right)\left( {{7^3} + {6^3}} \right) \cr & = \left( {343 - 216} \right) \times \left( {343 + 216} \right) \cr & = 127 \times 559 \cr & = 127 \times 13 \times 43 \cr} $$
Clearly, it is divisible by 127, 13 as well as 559
88. After the division of a number successively by 3, 4 and 7, the remainder obtained is 2, 1 and 4 respectively. What will be remainder if 84 divide the same number?
a) 80
b) 75
c) 42
d) 53
Discussion
Explanation: As the Number gives a remainder of 4 when it is divided by 7, then the number must be in form of (7x + 4)
The same gives remainder 1 when it is divided 4, so the number must be in the form of {4 × (7x + 4) + 1}
Also, the number when divided by 3 gives remainder 2, thus number must be in form of [3 × {4 × (7x + 4) + 1} + 2]
On simplifying,
[3 × {4 × (7x + 4) + 1} + 2]
= 84x + 53
We get the final number 53 more than a multiple of 84. Hence, if the number is divided by 84,
The remainder will be 53
89. Find the remainder when 2256 is divided by 17.
a) 1
b) 16
c) 14
d) None of these
Discussion
Explanation:
$$\eqalign{ & \frac{{{2^{256}}}}{{17}} \cr & {\text{We}}\,{\text{can}}\,{\text{write}}\,{\text{it}}\,{\text{as}}:\,{\left( {{2^4}} \right)^{64}} \cr & \frac{{{{16}^{64}}}}{{17}} \cr & {\text{Individually, when 16 is divided by 17,}} \cr & {\text{gives a negative reminder of - 1}}{\text{.}} \cr & {\text{Required Remainder}}, \cr & {\left( { - 1} \right)^{64}} = 1 \cr} $$ .
90. Find the remainder when 496 is divided by 6.
a) 0
b) 2
c) 3
d) 4
Discussion
Explanation: $$\frac{{{4^{96}}}}{6},$$ we can write it in this form
$$\frac{{{{\left( {6 - 2} \right)}^{96}}}}{6}$$
Now, Remainder will depend only the powers of -2.
$$\frac{{{{\left( { - 2} \right)}^{96}}}}{6},$$ it is same as
$$\frac{{{{\left( {{{\left[ { - 2} \right]}^4}} \right)}^{24}}}}{6},$$ it is same as
$$\frac{{{{\left( {16} \right)}^{24}}}}{6}$$
$$\frac{{\left( {16 \times 16 \times 16 \times 16{\kern 1pt} ......{\kern 1pt} 24{\kern 1pt} {\text{times}}} \right)}}{6}$$
On dividing individually 16 we always get a remainder 4.
$$\frac{{\left( {4 \times 4 \times 4 \times 4{\kern 1pt} ......{\kern 1pt} 24{\kern 1pt} {\text{times}}} \right)}}{6}$$
Required Remainder = 4
Note: When 4 has even number of powers, it will always give remainder 4 on dividing by 6.
91. Find the remainder when 73 × 75 × 78 × 57 × 197 × 37 is divided by 34.
a) 32
b) 30
c) 15
d) 28
Discussion
Explanation: Remainder,
$$\frac{{73 \times 75 \times 78 \times 57 \times 197 \times 37}}{{34}}$$
$$ = \frac{{5 \times 7 \times 10 \times 23 \times 27 \times 3}}{{34}}$$
[We have taken individual remainder, which means if 73 is divided by 34 individually, it will give remainder 5, 75 divided 34 gives remainder 7 and so on.]
$$\eqalign{ & \frac{{5 \times 7 \times 10 \times 23 \times 27 \times 3}}{{34}} \cr & = \frac{{35 \times 30 \times 23 \times 27}}{{34}} \cr & = \frac{{1 \times - 4 \times - 11 \times - 7}}{{34}} \cr} $$
[We have taken here negative as well as positive remainder at the same time. When 30 divided by 34 it will give either positive remainder 30 or negative remainder -4. We can use any one of negative or positive remainder at any time.]
$$\eqalign{ & = \frac{{28 \times - 11}}{{34}} \cr & = \frac{{ - 6 \times - 11}}{{34}} \cr & = \frac{{66}}{{34}} \cr & {\text{R}} = 32 \cr} $$
Required remainder = 32
92. Find the remainder when 6799 is divided by 7.
a) 4
b) 6
c) 1
d) 2
Discussion
Explanation:
$$\eqalign{ & {\text{Remainder of}}\frac{{{{67}^{99}}}}{7} \cr & R = \frac{{{{\left( {63 + 4} \right)}^{99}}}}{7} \cr} $$
63 is divisible by 7 for any power, so required remainder will depend on the power of 4
Required remainder
$$\eqalign{ & \frac{{{4^{99}}}}{7} = = R = = \frac{{{4^{\left( {96 + 3} \right)}}}}{7} \cr & \frac{{{4^3}}}{7} \Rightarrow \frac{{64}}{7} \Rightarrow \frac{{\left( {63 + 1} \right)}}{7} = = R \Rightarrow 1 \cr & {\text{Note}}: \cr & \frac{4}{7}{\text{remainder}} = 4 \cr & \frac{{\left( {4 \times 4} \right)}}{7} = \frac{{16}}{7}{\text{remainder}} = 2 \cr & \frac{{\left( {4 \times 4 \times 4} \right)}}{7} = \frac{{64}}{7} = 1 \cr & \frac{{\left( {4 \times 4 \times 4 \times 4} \right)}}{7} = \frac{{256}}{7}{\text{remainder}} = 4 \cr & \frac{{\left( {4 \times 4 \times 4 \times 4 \times 4} \right)}}{7} = 2 \cr} $$
If we check for more power we will find that the remainder start repeating themselves as 4, 2, 1, 4, 2, 1 and so on. So when we get A number having greater power and to be divided by the other number B, we will break power in (4n + x) and the final remainder will depend on x i.e. $$\frac{{{{\text{A}}^{\text{x}}}}}{{\text{B}}}$$
93. Let N = 1421 × 1423 × 1425. What is the remainder when N is divided by 12?
a) 0
b) 9
c) 3
d) 6
Discussion
Explanation: Remainder,
$$\eqalign{ & \frac{{1421 \times 1423 \times 1425}}{{12}} = R \cr & R \Rightarrow \frac{{5 \times 7 \times 9}}{{12}} \cr} $$
[Here, we have taken individual remainder such as 1421 divided by 12 gives remainder 5, 1423 and 1425 gives the remainder as 7 and 9 on dividing by 12.]
The sum is reduced to,
$$\frac{{5 \times 7 \times 9}}{{12}} = \frac{{35 \times 9}}{{12}}$$
$$\frac{{35 \times 9}}{{12}}$$ = Remainder ⇒ -1 × -3 = 3 [Here, we have taken negative remainder] So, required remainder will be 3.
Note: When, $$\frac{9}{{12}}$$ it gives positive remainder as 9 and it also give a negative remainder -3. As per our convenience,we can take any time positive or negative remainder.
94. Three numbers are in ratio 1 : 2 : 3 and HCF is 12. The numbers are:
a) 12, 24, 36
b) 11, 22, 33
c) 12, 24, 32
d) 5, 10, 15
Discussion
Explanation: Since, the numbers are given in the form of ratio that means their common factors have been cancelled.
Each one's common factor is HCF.
Here HCF = 12,
The numbers are 12, 24 and 36.
95. What is the least number of soldiers that can be drawn up in troops of 12, 15, 18 and 20 soldiers and also in form of a solid square?
a) 900
b) 400
c) 1600
d) 2500
Discussion
Explanation: We need to find out the LCM of the given numbers.
LCM of 12, 15, 18 and 20;
$$\eqalign{ & 12 = 2 \times 2 \times 3; \cr & 15 = 3 \times 5; \cr & 18 = 2 \times 3 \times 3; \cr & 20 = 2 \times 2 \times 5; \cr} $$
LCM = $$2 \times 2 \times 3 \times 5 \times 3$$
Since, the soldiers are in the form of a solid square.
Hence, LCM must be a perfect square. To make the LCM a perfect square, We have to multiply it by 5,
The required number of soldiers
= $$2 \times 2 \times 3 \times 3 \times 5 \times 5$$
= 900
96. Find the remainder when 65203 is divided by 7.
a) 4
b) 2
c) 1
d) 6
Discussion
Explanation:
$$\eqalign{ & \frac{{{{65}^{203}}}}{7} \cr & \frac{{{{\left( {63 + 2} \right)}^{203}}}}{7} \cr & 63\,{\text{is}}\,{\text{divisible}}\,{\text{by}}\,7, \cr & {\text{so}}\,{\text{remainder}}\,{\text{will}}\,{\text{depend}}\,{\text{on}}\,{\text{the}}\,{\text{powers}}\,{\text{of}}\,2 \cr & \frac{{{2^{203}}}}{7} \cr & {\text{Its}}\,{\text{remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}} \cr & \frac{{{2^3}}}{7} \cr & {\text{Required}}\,{\text{Remainder}}\, \cr & \frac{8}{7} = 1\cr & {\text{Required}}\,{\text{remainder}} = 1 \cr} $$
Note: We have manipulated the powers in the form of (4x + n). It means 203 is taken as,
203 = 4x + n = 4 × 50 + 3.
We neglect power which is in the multiple of 4.
97. Find the remainder when 67107 is divided by 7.
a) 4
b) 2
c) 1
d) 6
Discussion
Explanation:
$$\eqalign{ & \frac{{{{67}^{107}}}}{7} \cr & \frac{{{{\left( {7 \times 9 + 4} \right)}^{107}}}}{7} \cr & {\text{The}}\,{\text{remainder}}\,{\text{will}}\,{\text{be}}\,{\text{same}}\,{\text{as}} \cr & \frac{{{4^{107}}}}{7} \cr & \frac{{{4^3}}}{7} \cr & \frac{{64}}{7} \cr & {\text{Required}}\,{\text{Remainder}} = 1 \cr} $$
98. Find the remainder when 54124 is divided by 17.
a) 8
b) 13
c) 16
d) 9
Discussion
Explanation:
$$\eqalign{ & \frac{{{{54}^{124}}}}{{17}} \cr & \frac{{{{\left( {17 \times 3 + 3} \right)}^{124}}}}{{17}} \cr & {\text{The}}\,{\text{remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}}, \cr & \frac{{{3^{124}}}}{{17}} \cr & \frac{{{3^{4 \times 31}}}}{{17}} \cr & {\text{Remainder}}\,{\text{will}}\,{\text{be}}\,{\text{same}}\,{\text{as}}, \cr & \frac{{{3^4}}}{{17}} \cr & \frac{{81}}{{17}} \cr & \text{Remainder} = 13 \cr} $$
99. Find unit digit of product (173)45 × (152)77 × (777)999.
a) 4
b) 2
c) 8
d) 6
Discussion
Explanation : To find unit digit of a number or an Expression, We have to divide the number or expression by 10 and the remainder obtained by this operation would be the required unit digit.
$$\eqalign{ & \frac{{ {{{\left( {173} \right)}^{45}} \times {{\left( {152} \right)}^{77}} \times {{\left( {777} \right)}^{999}}} }}{{10}} \cr & {\text{Remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}}, \cr & \frac{{ {{3^{45}} \times {2^{77}} \times {7^{999}}} }}{{10}} \cr & \frac{{ {3 \times 2 \times {7^3}} }}{{10}} \cr & \frac{{ {6 \times 343} }}{{10}} \cr & {\text{Remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}}\,\frac{{ {6 \times 3} }}{{10}} \cr }$$
Required remainder and unit digit will be 8.
100. The square of a number greater than 1000 that is not divisible by three, when divided by three, leaves a remainder of
a) 1 always
b) 2 always
c) Either 1 or 2
d) 0
Discussion
Explanation: In such cases remainder will always be 1.