1. 0.1 and $$\frac{5}{8}$$ of a bamboo are in mud and water respectively and the rest length 2.75 m is above water. What is the length of the bamboo ?
a) 10 m
b) 30 m
c) 27.5 m
d) 20 m
Explanation: $${\text{0}}{\text{.1 = }}\frac{1}{{10}}\left( {{\text{mud}}} \right),\,\,\,\,\frac{5}{8}\left( {{\text{water}}} \right)$$
Let total length of bamboo = 40x ( LCM of 8, 10 )
$$\eqalign{ & {\text{Mud part :}} \cr & {\text{ = }}\frac{1}{{10}} \times 40x = 4x \cr & {\text{Water part :}} \cr & {\text{ = }}\frac{5}{8} \times 40x = 25x \cr & {\text{Remaining part :}} \cr & = (40x - 25x - 4x) \cr & = 11x \cr & 11x{\text{ }} \to {\text{ }}2.75m \cr & {\text{ }}x{\text{ }} \to {\text{ }}0.25m \cr & 40x{\text{ }} \to {\text{ }}10m \cr} $$
2. The last digit of (1001)2008 + (1002) = ?
a) 0
b) 3
c) 4
d) 6
Explanation:
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{(1001)^{2008}} + (1002) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \cr & {\text{unit digit}} \to {\text{ }}{{\text{1}}^{2008}} + 1002 \cr} $$
Unit digit will be 1 in case of 1 respective of power
⇒ 1 + 1002 = 1003
⇒ 3 unit digit (last digit)
3. The number 0.121212 . . . . . in the form $$\frac{p}{q}$$ is equal to
a) $$\frac{4}{11}$$
b) $$\frac{2}{11}$$
c) $$\frac{4}{33}$$
d) $$\frac{2}{33}$$
Explanation:
$$\eqalign{ & {\text{0}}{\text{.121212}}........ \cr & \Rightarrow 0.\overline {12} = \frac{{12}}{{99}} = \frac{4}{{33}} \cr} $$
4. By how much does $$\frac{6}{{\frac{7}{8}}}$$ exceed $$\frac{{\frac{6}{7}}}{8}$$ ?
a) $$6\frac{1}{8}$$
b) $$6\frac{3}{4}$$
c) $$7\frac{3}{4}$$
d) $$7\frac{5}{6}$$
Explanation:
$$\eqalign{ & \frac{6}{{\frac{7}{8}}} = \frac{{6 \times 8}}{7} = \frac{{48}}{7} \cr & \frac{{\frac{6}{7}}}{8} = \frac{6}{{7 \times 8}} = \frac{6}{{56}} = \frac{3}{{28}} \cr & {\text{Difference - }} \cr & = \frac{{48}}{7} - \frac{3}{{28}} \cr & = \frac{{192 - 3}}{{28}} \cr & = \frac{{189}}{{28}} \cr & = \frac{{27}}{4} \cr & = 6\frac{3}{4} \cr} $$
5. The simplified value of $$\left( {1 - \frac{1}{3}} \right)$$ $$\left( {1 - \frac{1}{4}} \right)$$ $$\left( {1 - \frac{1}{5}} \right)$$ . . . . .$$\left( {1 - \frac{1}{99}} \right)$$ $$\left( {1 - \frac{1}{100}} \right)$$
a) $$\frac{2}{{99}}$$
b) $$\frac{1}{{25}}$$
c) $$\frac{1}{{50}}$$
d) $$\frac{1}{{100}}$$
Explanation:
$$\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right)\left( {1 - \frac{1}{5}} \right)$$ . . . . .$$\left( {1 - \frac{1}{{99}}} \right)$$ $$\left( {1 - \frac{1}{{100}}} \right)$$
$$\eqalign{ & = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times ..... \times \frac{{98}}{{99}} \times \frac{{99}}{{100}} \cr & = \frac{2}{{100}} \cr & = \frac{1}{{50}} \cr} $$
6. The sum of three numbers is 2, the 1st number is $$\frac{1}{2}$$ times the 2nd and the 3rd number is $$\frac{1}{4}$$ times the 2nd number. The 2nd number is :
a) $$\frac{7}{6}$$
b) $$\frac{8}{7}$$
c) $$\frac{9}{8}$$
d) $$\frac{10}{9}$$
Explanation: Let the number is a, b, c
a + b + c = 2
a = $$\frac{1}{2}$$ b.....(i)
c = $$\frac{1}{4}$$ b.....(ii)
$$\frac{a}{b}$$ = $$\frac{1}{2}$$
$$\frac{b}{c}$$ = $$\frac{4}{1}$$
| a | : | b | : | c |
| 1 | : | 2 | ↘ | |
| ↘ | 4 | : | 1 | |
| 4 | : | 8 | : | 2 |
(2 + 4 + 1) = 7
7 units → 2
1 unit → $$\frac{2}{7}$$
4 units → $$\frac{2}{7}$$ × 4 = $$\frac{8}{7}$$
So, 2nd number is $$\frac{8}{7}$$
7. If the product of two positive numbers be 1575 and their ration is 7 : 9, then the greatest number is
a) 45
b) 135
c) 35
d) 63
Explanation: Let the numbers are 7x and 9x
According to the question,
$$\eqalign{ & \Rightarrow 7x \times 9x = 1575 \cr & \Rightarrow 63{x^2} = 1575 \cr & \Rightarrow {x^2} = 25 \cr & \Rightarrow x = 5 \cr} $$
Then greater number
= 9x
= 9 × 5
= 45
8. A number when divided by 729 given a remainder of 56. What will we get as remainder if the same number is divided by 27 ?
a) 4
b) 2
c) 0
d) 1
Explanation: If first number (729) is completely divide by second number (27) then we divide remainder from second number then remaining remainder will be answer.
$$\frac{729}{27}$$ = 27 times
Now, $$\frac{56}{27}$$ = 2 × 27 + 2 → Remainder
9. Sum of two numbers is thrice their difference. Their ratio is :
a) 1 : 2
b) 2 : 1
c) 3 : 1
d) 1 : 3
Explanation : Let the numbers = a, b
According to the question,
a + b = 3 (a - b)
$$\frac{a + b}{a - b}$$ = $$\frac{3}{1}$$
By componendo and dividendo
$$\frac{2a}{2b}$$ = $$\frac{4}{2}$$
$$\frac{a}{b}$$ = $$\frac{2}{1}$$
a : b = 2 : 1
10. A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as the remainder. The number is
a) 220030
b) 22030
c) 1220
d) 1250
Explanation: Let number is N
N is divided by the sum of 555 and 445, the quotient is 2 times difference
Quotient = 2 (555 - 445) = 220
Remainder = 30 (given)
N = 1000 × 220 + 30
= 220030
11. The real number to be added to 13851 to get a number which is divisible by 87 is :
a) 18
b) 43
c) 54
d) 69
Explanation: By hit and trial method
= 13851 + 69
= 13920
Which is divisible by 87
12. Among the following statements, the statement which is 'not correct' is :
a) Every natural number is a real number
b) Every real number is a rational number
c) Every integer is a rational number
d) Every natural number is an integer
Explanation: Every real number is a rational number
13. Sum of three consecutive integers is 51. The middle one is
a) 14
b) 15
c) 16
d) 17
Explanation: Let numbers are a, a + 1, a + 2
Then,
a + a + 1 + a + 2 = 51
3a + 3 = 51
3a = 48
a = 16
Middle number :
= a + 1
= 16 + 1
= 17
14. The sum of 10 terms of the arithmetic series is 390. If the third term of the series is 19. Find the first term :
a) 3
b) 5
c) 7
d) 8
Explanation: Sum of A.P. = $$\frac{n}{2}\left[ {2a + (n - 1)d} \right]$$
nth term = a + (n - 1)d
3rd term :
⇒ a + (3 - 1)d = 19
⇒ a + 2d = 19.....(i)
Sum of 10 term :
$$\eqalign{ & \Rightarrow \frac{{10}}{2}\left[ {2a + 9d} \right] = 390 \cr & \Rightarrow 2a + 9d = 78..... (ii) \cr & \Rightarrow a + 2d = 19 \cr} $$
From equation (i) and (ii)
a = 3
d = 8
15. If p = - 0.12, q = - 0.01 and r = - 0.015, then the correct relationship among the three is :
a) p < r < q
b) p > r > q
c) p < q < r
d) q > p > r
Explanation: p = - 0.12
q = - 0.01
r = - 0.015
If all values would be positive
0.12 > 0.015 > 0.01
p > r > q
But these are negative so their order will be
q > r > p
16. 12345679 × 72 is equal to
a) 88888888
b) 999999998
c) 888888888
d) 898989898
Explanation: 12345679 × 72
Shortcut method :
Since 72 is factor of 9, So the answer should be divisible by 9.
Now check with option by adding the digits of option and check which one is divisible by 9.
Here option (c) 888888888 is divisible by 9, so this is the multiple of 12345679 × 72
17. Which of the following fraction is greater than $$\frac{3}{4}$$ but less than $$\frac{5}{6}$$ ?
a) $$\frac{2}{3}$$
b) $$\frac{1}{2}$$
c) $$\frac{4}{5}$$
d) $$\frac{9}{10}$$
Explanation:
$$\eqalign{ & \frac{3}{4} = 75\% ,\frac{5}{6} = 83.33\% , \cr & (1)\frac{2}{3} = 66.66\% , \cr & (2)\frac{1}{2} = 50\% , \cr & (3)\frac{4}{5} = 80\% , \cr & (4)\frac{9}{{10}} = 90\% \cr & {\text{Hence , }}\frac{4}{5}{\text{ is between the fraction}}{\text{.}} \cr} $$
18. The digit in unit's place of the product 81 × 82 × 83 × ..... × 89 is :
a) 0
b) 2
c) 6
d) 8
Explanation: 81 × 82 × 83 ×..... × 89 take unit digit multiply
1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 = 0
Since there is 5 and 2 are present
So we get 5 × 2 = 10 and the last digit of 10 is 0, so the last digit of the above number is also 0
19. $$\left( {999\frac{{999}}{{1000}} \times 7} \right)$$
is equal to
a) $${\text{6993}}\frac{7}{{1000}}$$
b) $${\text{7000}}\frac{7}{{1000}}$$
c) $${\text{6633}}\frac{7}{{1000}}$$
d) $${\text{6999}}\frac{{993}}{{1000}}$$
Explanation :
$$\eqalign{ & = \left( {999\frac{{999}}{{1000}} \times 7} \right) \cr & = \left( {999\frac{{999}}{{1000}}} \right) \times 7 \cr & = 6993 + \frac{{6993}}{{1000}} \cr & = 6993 + 6\frac{{993}}{{1000}} \cr & = 6993 + 6 + \frac{{993}}{{1000}} \cr & = 6999\frac{{993}}{{1000}} \cr} $$
20. Number 2, 4, 6, 8, 10.....196, 198, 200 are multiplied together. The number of zeros at the end of the product on the right will be equal to
a) 21
b) 22
c) 24
d) 25
Explanation:
$$\eqalign{ & {\text{2, 4, 6, 8, 10}}......{\text{198, 200 }} \cr & {2^{100}}(1 \times 2 \times 3 \times ..... \times 99 \times 100) \cr} $$
$$\eqalign{ & \underline {5\,\,\,|\,\,100} \cr & \underline {5\,\,\,|\,\,\,20} \,\,\,\,\, = 24{\text{ zero's (20 + 4)}} \cr & \,\,\,\,\,|\,\,\,4 \cr} $$
When we multiply the series of 1 × 2 × 3 ×..... × 100 then we find 24 zero at the end of the product
21. When simplified the product : $$\left( {2 - \frac{1}{3}} \right)$$ $$\left( {2 - \frac{3}{5}} \right)$$ $$\left( {2 - \frac{5}{7}} \right)$$ .....$$\left( {2 - \frac{{997}}{{999}}} \right)$$
a) $$\frac{1001}{3}$$
b) $$\frac{5}{3}$$
c) $$\frac{5}{999}$$
d) $$\frac{1001}{999}$$
Explanation: According to the question,
$$ \Rightarrow \left( {2 - \frac{1}{3}} \right)\left( {2 - \frac{3}{5}} \right)\left( {2 - \frac{5}{7}} \right).....$$ $$\left( {2 - \frac{{997}}{{999}}} \right)$$
$$\eqalign{ & \Rightarrow \frac{5}{3} \times \frac{7}{5} \times \frac{9}{7}.....\frac{{1001}}{{999}} \cr & \Rightarrow \frac{1}{3} \times 1001 \cr & \Rightarrow \frac{{1001}}{3} \cr} $$
22. The smallest possible three placed decimal is
a) 0.012
b) 0.123
c) 0.111
d) None of these
Explanation: Smallest number in case of decimal = 0.001
23. The divisor is 25 times of the quotient and 5 times the remainder. If the quotient is 16, the dividend is
a) 6400
b) 6480
c) 400
d) 480
Explanation: Dividend = (divisor × quotient) + remainder
According to the question
$$\eqalign{ & {\text{Divisor :}} \cr & \Rightarrow {\text{16}} \times {\text{25 = 5}} \times {\text{remainder}} \cr & \Rightarrow {\text{R = }}\frac{1}{5} \times 16 \times 25 \cr & {\text{Dividend :}} \cr & {\text{ = }}\left[ {\left( {{\text{16}} \times {\text{25}}} \right) \times 16} \right] + \frac{1}{5} \times 16 \times 25 \cr & = \left[ {16 \times 25 \times 16} \right] + 80 \cr & = 6480 \cr} $$
24. A runner runs $$1\frac{1}{4}$$ laps of a 5 laps race. What laps of the race remains to be run ?
a) $$\frac{15}{4}$$
b) $$\frac{4}{5}$$
c) $$\frac{5}{6}$$
d) $$\frac{2}{3}$$
Explanation: Remaining distance to be run
$$\eqalign{ & = 5 - \frac{5}{4} \cr & = \frac{{20 - 5}}{4} \cr & = \frac{{15}}{4}\,{\text{laps}} \cr} $$
25. $$\frac{1}{{10}}$$ of a rod is coloured red, $$\frac{1}{{20}}$$ orange, $$\frac{1}{{30}}$$ yellow, $$\frac{1}{{40}}$$ green, $$\frac{1}{{50}}$$ blue, $$\frac{1}{{60}}$$ black and the rest is violet. If the length of the violet part of the rod is 12.08 mtr, then the length of the rod is
a) 16 m
b) 18 m
c) 20 m
d) 30 m
Explanation: Let the total length of rod is 1 unit
According to question,
$$ \Rightarrow 1 - \frac{1}{{10}} - \frac{1}{{20}} - \frac{1}{{30}} - \frac{1}{{40}} - \frac{1}{{50}}$$ $$ - \frac{1}{{60}}$$
$$ \Rightarrow \frac{{600 - 60 - 30 - 20 - 15 - 12 - 10}}{{600}}$$
$$\eqalign{ & \Rightarrow \frac{{453 \leftarrow {\text{ Violet part}}}}{{600 \leftarrow {\text{ Total part}}}} \cr & 453{\text{ part }} \to {\text{ 12}}{\text{.08 mtr}} \cr & {\text{1 part }} \to \,\,\,\,\,\,\,\,\frac{{12.08}}{{453}} \cr & 600{\text{ part }} \to \frac{{12.08}}{{453}} \times 600 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 16{\text{mtr}} \cr} $$
So total length of rod is 16 metres
26. The sum of two numbers is 75 and their difference is 25. The product of the two numbers is :
a) 1350
b) 1250
c) 125
d) 1000
Explanation: Let the numbers are a, b
⇒ a + b = 75.....(i)
⇒ a - b = 25.....(ii)
⇒ After solving (i) and (ii)
⇒ a = 50, b = 25
⇒ ab = 50 × 25 = 1250
So, their product is 1250
27. The reciprocals of the squares of the number $$1\frac{1}{2}$$ and $$1\frac{1}{3}$$ are in the ratio :
a) 64 : 81
b) 8 : 9
c) 81 : 64
d) 9 : 8
Explanation: $$1\frac{1}{2}$$ = $$\frac{3}{2}$$
$$1\frac{1}{3}$$ = $$\frac{4}{3}$$
Square = $$\frac{9}{4}$$ $$\frac{16}{9}$$
Reciprocal ratio
$$\frac{4}{9}$$ : $$\frac{9}{16}$$
64 : 81
28. Which is the largest among the numbers : $$\root 3 \of 7 ,\root 4 \of {13} ,\sqrt 5 $$
a) $$\sqrt 5 $$
b) $$\root 3 \of 7 $$
c) $$\root 4 \of {13} $$
d) All are equal
Explanation:
$$\root 3 \of 7 ,\root 4 \of {13} ,\sqrt 5 $$
Taking L.C.M. of 3, 4 and 2.
| $${7^{\frac{1}{3} \times 12}}$$ | $${13^{\frac{1}{4} \times 12}}$$ | $${5^{\frac{1}{2} \times 12}}$$ |
| 74 | 133 | 56 |
| ↓ | ↓ | ↓ |
| 2401 | 2197 | 15625 |
29. When a number is divided by 5, the remainder is 3. What will be the remainder when sum of cube of that number and square of that number is divided by 5 ?
a) 1
b) 2
c) 3
d) 4
Explanation: Let us assume any such number which when divided by 5 leaves remainder as 3.
Let it be 8
So, now
$$\eqalign{ & = \frac{{{{\left( 8 \right)}^2} + {{\left( 8 \right)}^3}}}{5} \cr & = \frac{{64 + 512}}{5} \cr & = \frac{{576}}{5} = 1\,\,\left[ {{\text{Remainder}}} \right] \cr} $$
30. The simplified value of $$\frac{{\left( {0.0539 - 0.002} \right) \times 0.4 + 0.56 \times 0.07}}{{0.04 \times 0.25}}$$
a) 599.6
b) 0.5996
c) 5.996
d) 59.96
Explanation:
$$\eqalign{ & = \frac{{\left( {0.0539 - 0.002} \right) \times 0.4 + 0.56 \times 0.07}}{{0.04 \times 0.25}} \cr & = \frac{{\left( {0.0519} \right) \times 0.4 + 0.56 \times 0.07}}{{0.04 \times 0.25}} \cr & = \frac{{0.02076 + 0.0392}}{{0.04 \times 0.25}} \cr & = \frac{{0.05996}}{{0.01}} \cr & = 5.996 \cr} $$
31. Which of the following is a perfect square ?
a) 497497
b) 4587632
c) 1046529
d) 1034758
Explanation: By checking last 3 digits
Then, last '3' digits of 1046529
529 is a square of 23
Hence, this is correct option
32. While solving a problem, by mistake, Anita squared a number and then subtracted 25 from it rather than first subtracting 25 from the number and then squaring it. But she got the right answer. What was the given number ?
a) 48
b) Cannot be determined
c) 13
d) 38
Explanation: Let the given number be x
According to the question,
x2 - 25 = (x - 25)2
x2 - 25 = x2 + (25)2 - 50x
x = 13
33. The decimal fraction $$2.3\overline {49} $$ is equal to :
a) $$\frac{{2326}}{{999}}$$
b) $$\frac{{2326}}{{990}}$$
c) $$\frac{{2347}}{{999}}$$
d) $$\frac{{2347}}{{990}}$$
Explanation: $$2.3\overline {49} = \frac{{2349 - 23}}{{990}} = \frac{{2326}}{{990}}$$
34. A number x when divided by 289 leave 18 as the remainder. The same number when divided by 17 leaves y as a remainder. the value of y is :
a) 5
b) 2
c) 3
d) 1
Explanation:
$$\eqalign{ & \frac{{{\text{Remainder of number}}}}{{17}} = \frac{{18}}{{17}} \cr & \Rightarrow {\text{Remainder = 1}} \cr} $$
35. The digit at Hundred's place value of 17! is :
a) 1
b) 0
c) 2
d) 3
Explanation: We know that after 5! we get one zero at the end of the number
And after 10! → two zeros
And after 15! → three zeros
We can say that in 17! we get minimum three zeros
The Hundred's place value of 17! = 0
36. Arrangement of the fractions into ascending order : $$\frac{4}{3},$$ $$ - \frac{2}{9},$$ $$ - \frac{7}{8},$$ $$\frac{5}{{12}}$$
a) $$ - \frac{2}{9}, - \frac{7}{8},\frac{4}{3},\frac{5}{{12}}$$
b) $$ - \frac{7}{8}, - \frac{2}{9},\frac{5}{{12}}\frac{4}{3}$$
c) $$ - \frac{7}{8}, - \frac{2}{9},\frac{4}{3},\frac{5}{{12}}$$
d) $$ - \frac{2}{9}, - \frac{7}{8},\frac{5}{{12}}\frac{4}{3}$$
Explanation:
$$\frac{4}{3}, - \frac{2}{9}, - \frac{7}{8},\frac{5}{{12}}$$
$$\eqalign{ & \frac{4}{3} = 1.33, \cr & - \frac{2}{9} = - 0.22, \cr & - \frac{7}{8} = - 0.875, \cr & \frac{5}{{12}} = 0.416 \cr} $$
So, $$ - \frac{7}{8}, - \frac{2}{9},\frac{5}{{12}},\frac{4}{3}$$
37. If 13 + 23 + .....+ 103 = 3025, then the value of 23 + 43 +.....+ 203 is :
a) 7590
b) 5060
c) 24200
d) 12100
Explanation: 13 + 23 + .....+ 103 = 3025
To find = 23 + 43 + 63 +.....+ 203 = ?
⇒ 23 (13 + 23 + .....+ 103)
⇒ 8 × 3025
⇒ 24200
38. There are 50 boxes and 50 persons. Person 1 keeps 1 marble in every box, person 2 keeps 2 marbles in every 2nd box, person 3 keeps 3 marbles in every third box. The process goes on till person 50 keeps 50 marbles in the 50th box. Find the total number of marbles kept in the 50th box.
a) 43
b) 78
c) 6
d) 93
Explanation: Marbles in the 50th box will be kept by 1st, 2nd, 5th, 10th, 25th and 50th person is a factor of 50.
Number of marbles
=1 + 2 + 5 + 10 + 25 + 50
= 93
39. The difference between the greatest and the least four digit numbers that begins with 3 and ends with 5 is :
a) 900
b) 999
c) 909
d) 990
Explanation: ⇒ The least number = 3005
⇒ The greatest number = 3995
⇒ Difference will be = 3995 - 3005 = 990
40. If in a three digits number the last two digits places are interchanged a new number is formed which is greater than the original number by 45. What is the difference between the last two digits of that number ?
a) 9
b) 8
c) 6
d) 5
Explanation: Three digit number = 100x + 10y + z
To make number after changing last two digit = 100x + 10z + y
Now,
100x + 10y + z = 100x + 10z + y - 45
9z - 9y = 45
z - y = 5
41. The product of two positive integers is 2048 and one of them is twice the other. Then the small of the number is :
a) 32
b) 64
c) 16
d) 1024
Explanation: Let first number is x and second number is 2x.
Then, according to the question
x × 2x = 2048
x2 = 1024
x = 32
Smaller number = 32
42. If $$\frac{{51.84}}{{4.32}} = 12$$ , then the value of $$\frac{{0.005184}}{{0.432}}$$ is :
a) 0.12
b) 0.012
c) 0.0012
d) 1.2
Explanation: Given :
$$\eqalign{ & \Leftrightarrow \frac{{51.84}}{{4.32}} = 12 \cr & \Leftrightarrow \frac{{5184}}{{432}} = 12 \cr & \therefore \frac{{0.005184}}{{0.432}} \cr & = \frac{{5.184}}{{432}} \cr & = \frac{{5184}}{{432}} \times \frac{1}{{1000}} \cr & = 12 \times \frac{1}{{1000}} \cr & = 0.012 \cr} $$
43. How many $$\frac{1}{6}$$ of together make $${\text{41}}\frac{2}{3}$$ ?
a) 125
b) 150
c) 250
d) 350
Explanation: $$\frac{{41\frac{2}{3}}}{{\frac{1}{6}}} = \frac{{125}}{3} \times \frac{6}{1} = 250{\text{ times}}$$
44. A girl was asked to multiply a number by $$\frac{7}{8}$$, instead she divided the number by $$\frac{7}{8}$$ and got the result 15 more than the correct result. The sum of the digits of the number was
a) 4
b) 8
c) 6
d) 11
Explanation:
$$\eqalign{ & {\text{Let the number be x}} \cr & {\text{According to question}} \cr & \frac{x}{{7/8}} - \frac{7}{8}x = 15 \cr & \frac{8}{7}x - \frac{7}{8}x = 15 \cr & \frac{{64x - 49x}}{{56}} = 15 \cr & 15x = 15 \times 56 \cr & x = 56 \cr & {\text{Sum of digits }} \cr & = 5{\text{ }} + {\text{ }}6 \cr & = 11 \cr} $$
45. The sum of three consecutive natural numbers each divisible by 5, is 225. The largest among them is ?
a) 85
b) 75
c) 70
d) 80
Explanation:
$$\eqalign{ & {\text{Let number are }}x,{\text{ }}x + 5,{\text{ }}x + 10 \cr & {\text{So, }}x{\text{ + }}x + 5 + x + 10 = 225 \cr & 3x + 15 = 225 \cr & 3x = 210 \cr & x = 70 \cr & {\text{largest number is }}70 + 10 = 80 \cr} $$
46. The sum of squares of three positive integers is 323. If the sum of squares of two numbers is twice the third, their product is :
a) 255
b) 260
c) 265
d) 270
Explanation: Let the integers in a, b, c
a2 + b 2 + c2 = 323.....(i)
a2 + b2 + = 2c.....(ii)
Break the 323 in to their squares
a = 5
b = 3
c = 17
So, a2 + b2 + c2 = 323
and
a2 + b2 = 2c
252 + 32 = 2 × 17
34 = 34
So, Satisfied
a : b : c = 17 × 5 × 3 = 255
47. Each member of a club contributes as much rupees and as much paise as the number of members of the club. If the total contribution is Rs. 2525, then the number of members of the club is :
a) 60
b) 45
c) 55
d) 50
Explanation: Let the total member be x
x members contribution is Rs. x
And each member also contributes x paise
So, total rupees is
x member Rs. x = Rs. x2
x member Rs. x paise = $$\frac{x.x}{100}$$
Total rupees is
$${x^2} + \frac{{{x^2}}}{{100}} = 2525$$
101x2 = 2525 × 100
101x2 = 101 × 25 × 100
x = 50
48. The sum of the cubes of two numbers in the ratio 3 : 4 is 5824. The sum of the numbers is :
a) $${\left( {5824} \right)^{\frac{1}{3}}}$$
b) 28
c) 24
d) 14
Explanation: Let numbers are = 3x, 4x
(3x)3 + (4x)3 = 5824
27x3 + 64x3 = 5824
91x3 = 5824
x3 = $$\frac{5824}{91}$$
x3 = 64
x = 4
So, numbers are
3x = 3 × 4 = 12
4x = 4 × 4 = 16
Sum = 12 + 16 = 28
49. The value of $$\left( {0.34\overline {67} + 0.13\overline {33} } \right)$$ is :
a) 0.48
b) $$0.48\overline {01} $$
c) $$0.\overline {48} $$
d) $$0.4\overline {8} $$
Explanation: 0.34676767 ..... + 0.13333333 .....
= 0.480101 .....
= $$0.48\overline {01} $$
50. Arrange the following fraction in decreasing order : $$\frac{3}{5},$$ $$\frac{7}{9}$$ $$,\frac{{11}}{{13}}$$
a) $$\frac{3}{5},\frac{7}{9},\frac{{11}}{{13}}$$
b) $$\frac{7}{9},\frac{3}{5},\frac{{11}}{{13}}$$
c) $$\frac{{11}}{{13}},\frac{7}{9},\frac{3}{5}$$
d) $$\frac{{11}}{{13}},\frac{3}{5},\frac{7}{9}$$
Explanation:
$$\frac{3}{5},\frac{7}{9},\frac{{11}}{{13}}$$
$$\eqalign{ & \frac{3}{5} = 0.60, \cr & \frac{7}{9} = 0.7, \cr & \frac{{11}}{{13}} = 0.8 \cr & \frac{{11}}{{13}}{\text{ > }}\frac{7}{9} > \frac{3}{5} \cr} $$
51. $$\left( {{3^{25}} + {3^{26}} + {3^{27}} + {3^{28}}} \right)$$ is divisible by :
a) 11
b) 16
c) 25
d) 30
Explanation:
$$\eqalign{ & \left( {{3^{25}} + {3^{26}} + {3^{27}} + {3^{28}}} \right)\, \cr & = {3^{25}}\left( {{3^0} + {3^1} + {3^2} + {3^3}} \right)\, \cr & = {3^{25}} \times 40 \cr & = {3^{24}} \times 120 \cr & {\text{now check with option }} \cr & {\text{Only 30 can divide this}}{\text{.}} \cr} $$
52. $${\text{999}}\frac{{998}}{{999}} \times 999$$ is equal to :
a) 998999
b) 999899
c) 989999
d) 999989
Explanation:
$$\eqalign{ & \Rightarrow {\text{999}}\frac{{998}}{{999}} \times 999 \cr & \Rightarrow \left( {999 + \frac{{998}}{{999}}} \right) \times 999 \cr & \Rightarrow \left[ {\left( {1000 - 1} \right) + \frac{{998}}{{999}}} \right] \times 999 \cr & \Rightarrow 999000 - 999 + 998 \cr & \Rightarrow 998999 \cr} $$
53. $$0.\overline {001} $$ is equal to
a) $$\frac{1}{1000}$$
b) $$\frac{1}{999}$$
c) $$\frac{1}{99}$$
d) $$\frac{1}{9}$$
Explanation: $$0.\overline {001} \,\, \Rightarrow \frac{1}{{999}}$$
54. If the sum of two numbers is 3 and the sum of their squares is 12, then their product is equal to
a) $$\frac{3}{2}$$
b) $$\frac{2}{3}$$
c) - $$\frac{3}{2}$$
d) - $$\frac{2}{3}$$
Explanation:
$$\eqalign{ & {\text{a + b = 3 }}.......{\text{ (i)}} \cr & {\text{(a and b are two numbers)}} \cr & {{\text{a}}^2}{\text{ + }}{{\text{b}}^2} = 12 \cr & {\text{On squaring}}\,\,{\text{equation (i)}} \cr & {\left( {{\text{a + b}}} \right)^2} = \,{3^2} \cr & {{\text{a}}^2}{\text{ + }}{{\text{b}}^2} + 2{\text{ab}} = 9 \cr & 12 + 2{\text{ab = 9}} \cr & {\text{2ab = - 3}} \cr & {\text{ab = }}-\frac{{3}}{2} \cr} $$
55. A man spends $$\frac{1}{3}$$ of his income on food, $$\frac{2}{5}$$ of his income on house rent, $$\frac{1}{5}$$ of his income on clothes. If he still has Rs. 400 left with him, his income is
a) Rs. 4000
b) Rs. 5000
c) Rs. 6000
d) Rs. 7000
Explanation: Let total income is 15x (L.C.M. of 3, 5)
$$\eqalign{ & {\text{Spend on food}} \cr & {\text{ = }}\frac{1}{3} \times 15x \cr & = 5x \cr & {\text{Spend on rent}} \cr & {\text{ = }}\frac{2}{5} \times 15x \cr & = 6x \cr & {\text{Spend on clothes }} \cr & {\text{ = }}\frac{1}{5} \times 15x \cr & = 3x \cr & {\text{Income left }} \cr & = 15x - \left( {5 + 6 + 3} \right)x \cr & = 15x - 14x = x \cr & x = 400 \cr & 15x = 6000 \cr & {\text{Income = 6000 }} \cr} $$
56. Which one of the following numbers is not a square of any natural number ?
a) 17956
b) 18225
c) 63592
d) 53361
Explanation: 63592 is not a square of any natural number
∴ Any number that have 2, 3, 7, 8 on its units place , It could not be perfect square of any number.
57. The fractions $$\frac{1}{3}{\text{,}}\frac{4}{7}$$ and $$\frac{2}{5}$$ written in ascending order given by
a) $$\frac{4}{7} < \frac{1}{3} < \frac{2}{5}$$
b) $$\frac{2}{5} < \frac{4}{7} < \frac{1}{3}$$
c) $$\frac{1}{3} < \frac{2}{5} < \frac{4}{7}$$
d) $$\frac{4}{7} < \frac{1}{3} < \frac{2}{5}$$
Explanation:
$$\eqalign{ & \frac{1}{3}{\text{,}}\frac{4}{7},\frac{2}{5} \cr & \frac{1}{3} = 0.33,{\text{ }} \cr & \frac{4}{7} = 0.5, \cr & \frac{2}{5} = 0.4 \cr & \frac{1}{3} < \frac{2}{5} < \frac{4}{7} \cr} $$
58. If 17200 is divided by 18, the remainder is
a) 17
b) 16
c) 1
d) 2
Explanation:
$$\eqalign{ & = {17^{200}} \div 18 \cr & = {\left( {18 - 1} \right)^{200}} \div 18 \cr} $$
Apply Binomial theorem
$$ = {\left( {18} \right)^{200}}{\left( { - 1} \right)^0} + {\left( {18} \right)^{199}}{\left( { - 1} \right)^1} + $$ $$..... + \, {\left( {18} \right)^1}{\left( { - 1} \right)^{199}}$$ $$ \, + \, {\left( {18} \right)^0}{\left( { - 1} \right)^{200}}$$
$$ \Rightarrow $$ Remainder always comes from last term is (18)0 (- 1)200
$$\eqalign{ & = \frac{{{{\left( {18} \right)}^0}{{\left( { - 1} \right)}^{200}}}}{{18}} \cr & = 1 \times {1^{200}} \cr & = 1 \cr & {\text{So, remainder}} = 1 \cr} $$
59. Given that, three numbers are such that the second number is twice the first and thrice the third. Also the average of the three numbers is 44. then the difference of the first and the third is :
a) 10
b) 11
c) 12
d) 13
Explanation: Let numbers are 3x, 6x, 2x
Average : $$\frac{11x}{3}$$ = 44
x = 12 = Difference
60. A number is doubled and 9 is added, If the resultant is tripled, it becomes 75. What is the number ?
a) 6
b) 3.5
c) 8
d) None of these
Explanation:
$$\eqalign{ & {\text{Let the number is }}x{\text{}} \cr & {\text{According to question,}} \cr & {\text{3}}\left( {2x + 9} \right) = 75 \cr & 2x + 9 = 25 \cr & x = \frac{{16}}{2} \cr & x = 8 \cr} $$
61. The unit digit in the product 771 × 663 × 365 = ?
a) 1
b) 2
c) 3
d) 4
Explanation:
| 771 | × | 663 | × | 365 | ||
| ↓ | ↓ | ↓ | ||||
| Unit Place | 73 | 63 | 31 | |||
| ↓ | ↓ | ↓ | ||||
| Unit Digit | ⇒ | 3 | × | 6 | × | 3 = 54 |
| ⇒ 4 is answer | ||||||
62. If sum of the two number is 80 and ratio is 3 : 5, then find numbers :
a) 50, 30
b) 60, 20
c) 20, 60
d) 30, 50
Explanation: Let number are a, b = 3x, 5x
So, 3x + 5x = 8x = 80
x = 10
So, number is :
3 × 10 = 30
5 × 10 = 50
63. $$\frac{2}{3}$$ of three-fourth of a number is :
a) $$\frac{1}{2}$$ of the number
b) $$\frac{1}{3}$$ of the number
c) $$\frac{8}{9}$$ of the number
d) $$\frac{11}{12}$$ of the number
Explanation:
$$\eqalign{ & = \frac{2}{3} \times \frac{3}{4} \times n\,\,\,\left( {{\text{n}}\,{\text{is}}\,{\text{a}}\,{\text{number}}} \right) \cr & = \frac{1}{2}n \cr} $$
64. The greatest four digit number which is exactly divisible by each one of the numbers 12, 18, 21 and 28 :
a) 9828
b) 9882
c) 9928
d) 9288
Explanation: L.C.M of 12, 18, 21, 28 = 252
9999 is divided by 252
Then remainder is 171
= 9999 - 171
The number will be = 9828
65. A number exceeds its two fifth by 75. The number is
a) 125
b) 100
c) 112
d) 150
Explanation: Let the number = x
⇒ According to the question,
$$\eqalign{ & \Rightarrow x - \frac{2}{5}x = 75 \cr & \Rightarrow \frac{{5x - 2x}}{5} = 75 \cr & \Rightarrow 3x = 75 \times 5 \cr & \Rightarrow x = 125 \cr} $$
66. A number of boys raised Rs. 400 for a famine relief fund, each boy giving as many 25 paise coins as there were boys. The number of boys was
a) 40
b) 16
c) 20
d) 100
Explanation:
$$\eqalign{ & {\text{Let number of boys}} = x \cr & {\text{Number of 25 paise coins}} = {x^2} \cr & {\text{According to question,}} \cr & \frac{{{x^2}}}{4} = 400 \cr & {x^2} = 1600 \cr & x = 40\,\, \cr} $$
67. The greatest whole number, by which the expression n4 + 6n3 + 11n2 + 6n + 24 is divisible for every natural number n , is
a) 6
b) 24
c) 12
d) 48
Explanation:
$$\eqalign{ & {\text{According to the question,}} \cr & {{\text{n}}^4} + 6{{\text{n}}^3} + 11{{\text{n}}^2} + 6{\text{n + 24}} \cr & {\text{put n = 1}} \cr & = 1 + 6 + 11 + 6 + 24 \cr & = 48 \cr & {\text{put n = 2}} \cr & {\text{ = 16 + 48 + 44 + 12 + 24}} \cr & {\text{ = 144}} \cr & {\text{Clearly it is divisible by 48}} \cr} $$
68. The sum of all natural numbers from 75 to 97 is
a) 1598
b) 1798
c) 1958
d) 1978
Explanation: The sum of all natural numbers upto 97 - sum of all natural numbers upto 74
sum of n natural numbers
$$\eqalign{ & = \frac{{n\left( {n + 1} \right)}}{2}\,\,\left( {{\text{Formula}}} \right) \cr & = \frac{{97 \times 98}}{2} - \frac{{74 \times 75}}{2} \cr & = 4753 - 2775 \cr & = 1978 \cr} $$
69. If $$\frac{4}{5}$$ of an estate be worth Rs. 16800, then the value of $$\frac{3}{7}$$ of it is
a) Rs. 90000
b) Rs. 9000
c) Rs. 72000
d) Rs. 21000
Explanation:
$$\eqalign{ & \frac{4}{5}{\text{ of an estate = 16800}} \cr & {\text{1 of an estate = 16800}} \times \frac{5}{4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 21000 \cr & \frac{3}{7}{\text{ of an estate = 21000}} \times \frac{3}{7} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 9000 \cr} $$
70. Which of the following number is not divided by 18 ?
a) 54036
b) 50436
c) 34056
d) 65043
Explanation: A number will divisible by 18 if it is divisible by 2 and 9
Clearly we can see 65043 is not divisible by 2
Because unit digit of 65043 is 3 so this will not be divisible by 18
71. The smallest number, which should be added to 756896, so as to obtain a multiple of 11, is :
a) 1
b) 2
c) 3
d) 5
Explanation: For divisibility of 11
= 756896
= (9 + 6 + 7) - (6 + 8 + 5)
= 3
So, 3 should be added to number to make it divisible by 11
72. If sum of two numbers be a and their product be b, then the sum of their reciprocals is :
a) $$\frac{{1}}{{b}}$$ + $$\frac{{1}}{{b}}$$
b) $$\frac{{b}}{{a}}$$
c) $$\frac{{a}}{{b}}$$
d) $$\frac{{1}}{{ab}}$$
Explanation: Let the two number are P and Q
P + Q = a
PQ = b
⇒ $$\frac{{1}}{{P}}$$ + $$\frac{{1}}{{Q}}$$
⇒ $$\frac{{Q + P}}{{PQ}}$$
⇒ $$\frac{{a}}{{b}}$$
73. If the numbers $$ \root 3 \of {9} , \root 4 \of 20 , \root 6 \of {25} $$ are arranged in ascending order, then the right arrangement is -
a) $$ \root 6 \of {25} , < \root 4 \of {20} , < \root 3 \of 9 $$
b) $$ \root 3 \of 9 , < \root 4 \of {20} , < \root 6 \of {25} $$
c) $$ \root 6 \of {25} , < \root 3 \of 9 , < \root 4 \of {20} $$
d) $$ \root 4 \of {20} , < \root 3 \of 9 , < \root 6 \of {25} $$
Explanation: $$\root 3 \of 9 ,\,\,\root 4 \of {20} ,\,\,\root 6 \of {25} $$
Taking L.C.M of 3, 4 and 6
$$\eqalign{ & = \root {12} \of {{9^4}} ,\,\,\root {12} \of {{{20}^3}} ,\,\,\root {12} \of {{{25}^2}} \cr & = \root {12} \of {6561} ,\,\,\root {12} \of {8000} ,\,\,\root {12} \of {625} \cr} $$
So, ascending order
$$ = \root 6 \of {25} , < \root 3 \of 9 , < \root 4 \of {20} $$
74. Which of the following fraction is the smallest ?
$$\frac{8}{{15}},$$ $$\frac{{14}}{{33}},$$ $$\frac{7}{{13}},$$ $$\frac{{11}}{{13}}$$
a) $$\frac{{8}}{{15}}$$
b) $$\frac{{14}}{{33}}$$
c) $$\frac{{7}}{{13}}$$
d) $$\frac{{11}}{{13}}$$
Explanation: $$\frac{8}{{15}},\frac{{14}}{{33}},\frac{7}{{13}},\frac{{11}}{{13}}$$
| (264) | , | (210) | |
| ↑ | ↑ | ||
| $$\frac{8}{{15}}$$ | ⤩ | $$\frac{14}{{33}}$$ | = $$\frac{8}{{15}}$$ > $$\frac{14}{{33}}$$ |
| (182) | , | (231) | |
| ↑ | ↑ | ||
| $$\frac{14}{{33}}$$ | ⤩ | $$\frac{7}{{13}}$$ | = $$\frac{14}{{33}}$$ < $$\frac{7}{{13}}$$ < $$\frac{11}{{13}}$$ |
$${\text{So}}\frac{{14}}{{33}}\,{\text{is}}\,{\text{smallest}}$$
75. When 'n' is divisible by 5 the remainder is 2. What is the remainder when n2 is divided by 5 ?
a) 2
b) 3
c) 1
d) 4
Explanation:
$$\eqalign{ & \frac{{\text{n}}}{5} \Rightarrow {\text{remainder}}\,\,2 \cr & {\text{if we put n = 7}} \cr} $$
Then it satisfies above situation
$$\eqalign{ & {\text{so n = 7}} \cr & \frac{{{{\text{n}}^2}}}{5} = \frac{{{7^2}}}{5} = \frac{{49}}{5} \cr & \Rightarrow {\text{remainder = 4}} \cr} $$
76. Sum of three consecutive even integers is 54. Find the least integer among them.
a) 18
b) 15
c) 14
d) 16
Explanation:
$$\eqalign{ & x + \left( {x + 2} \right) + \left( {x + 4} \right) = 54 \cr & 3x + 6 = 54 \cr & 3x = 48 \cr & x = 16 \cr & \cr} $$
77. On multiplying a number by 7 all the digit in the product appear as 3's , the smallest such numbers is
a) 47649
b) 47719
c) 47619
d) 48619
Explanation:
$$\eqalign{ & {\text{According to question,}} \cr & x \times 7 = 333333 \cr} $$
(In answer 5 digit number are given, so we take 6 digit 333333)
$$\eqalign{ & x = \frac{{333333}}{7} \cr & x = 47619 \cr} $$
78. Find the largest number, which exactly divides every number of the form (n3 - n) (n - 2) where n is a natural number greater than 2.
a) 6
b) 12
c) 24
d) 48
Explanation:
$$\eqalign{ & \Rightarrow \left( {{n^3} - n} \right)\left( {n - 2} \right){\text{put n = 3}} \cr & \Rightarrow \left( {{3^3} - 3} \right)\left( {3 - 2} \right) \cr & \Rightarrow \left( {27 - 3} \right) \times 1 \cr & \Rightarrow 24 \cr & {\text{It is divisible by }}24 \cr} $$
79. Two number when divided by 17, leaves remainder 13 and 11 respectively. If the sum of those two numbers is divided by 17, the remainder will be ?
a) 13
b) 11
c) 7
d) 4
Explanation: Dividend = divisor × quotient + remainder
$$\eqalign{ & {\text{First number }} \cr & \Rightarrow {\text{ }}\left( {17 \times {\text{n}}} \right) + 13 \cr & {\text{Let n = 1}} \cr & \Rightarrow (17 \times 1) + 13 \cr & \Rightarrow 30 \cr & {\text{second number }} \cr & \Rightarrow {\text{ }}\left( {17 \times {\text{n}}} \right) + 11 \cr & \Rightarrow (17 \times 1) + 11 \cr & \Rightarrow 28 \cr & {\text{According to question}} \cr & = \frac{{30 + 28}}{{17}} \cr & = \frac{{58}}{{17}} \cr & \Rightarrow {\text{remainder = 7}} \cr} $$
80. Unit digit in $${\left( {264} \right)^{102}} + {\left( {264} \right)^{103}}$$ is :
a) 0
b) 4
c) 6
d) 8
Explanation:
$${\left( {264} \right)^{102}} + {\left( {264} \right)^{103}}$$
unit digit
$$\eqalign{ & {{\text{4}}^1} \to 4 \to 4 \cr & {4^2} \to 16 \to 6 \cr & {4^3} \to 64 \to 4 \cr} $$
Rule: When 4 has odd power, then unit digit is 4
When 4 has even power, then unit digit is 6
$$\eqalign{ & {\left( {264} \right)^{102}} + {\left( {264} \right)^{103}} \cr & \,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \cr & \,\,\,\,{4^{102}}\,\,\,\,\,\,\, + \,\,\,\,\,\,{4^{103}} \cr} $$
$${\text{unit digit}} = 6 + 4 = 10 \to 0$$
(even power) (odd power)
81. (12)3 × 64 × 432-1 = ?
a) 5184
b) 5060
c) 5148
d) None of these
Explanation:
$$\eqalign{ & {\text{Given}}\,{\text{Exp}}{\text{.}} \cr & = \frac{{{{\left( {12} \right)}^3} \times {6^4}}}{{432}} \cr & = \frac{{{{\left( {12} \right)}^3} \times {6^4}}}{{12 \times {6^2}}} \cr & = {\left( {12} \right)^2} \times {6^2} \cr & = {\left( {72} \right)^2} \cr & = 5184 \cr} $$
82. 72519 × 9999 = ?
a) 725117481
b) 674217481
c) 685126481
d) 696217481
Explanation: 2519 × 9999
= 72519 × (10000 - 1)
= 72519 × 10000 - 72519 × 1
= 725190000 - 72519
= 725117481
83. Which of the following number is divisible by 24 ?
a) 35718
b) 63810
c) 537804
d) 3125736
Explanation: 24 = 3 × 8, where 3 and 8 co-prime
Clearly, 35718 is not divisible by 8, as 718 is not divisible by 8
Similarly, 63810 is not divisible by 8 and 537804 is not divisible by 8
Consider option (D),
Sum of digits = (3 + 1 + 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 3
Also, 736 is divisible by 8
∴ 3125736 is divisible by (3 × 8), i.e., 24
84. On dividing a number by 5, we get 3 as remainder. What will the remainder when the square of the this number is divided by 5 ?
a) 0
b) 1
c) 2
d) 4
Explanation: Let the number be x and on dividing x by 5, we get k as quotient and 3 as remainder
∴ x = 5k + 3
⇒ x2 = (5k + 3)2
⇒ x2 = (25k2 + 30k + 9)
⇒ x2 = 5 (5k2 + 6k + 1) + 4
∴ On dividing x2 by 5, we get 4 as remainder
85. On dividing a number by 357, we get 39 as remainder. On dividing the same number 17, what will be the remainder ?
a) 0
b) 3
c) 5
d) 11
Explanation: Let x be the number and y be the quotient. Then,
x = 357 × y + 39
= (17 × 21 × y) + (17 × 2) + 5
= 17 × (21y + 2) + 5
∴ Required remainder = 5
86.What least value must be assigned to '*' so that the number 63576 * 2 is divisible by 8 ?
a) 1
b) 2
c) 3
d) 4
Explanation: For divisibility by 8, last 3 number should be divisible by 8.
Now, put the value 0, 1 ... 2 ... and check.
When use * = $$\frac{632}{8}$$ = 79
So, hence it is divisible by 8 answer is 3
87. The number 1, 3, 5, 7. . . . . 99 and 128 are multiplied together. The number of zeros at the end of the product must be
a) 19
b) 22
c) 7
d) Nil
Explanation: →( 1, 3, 5, 7. . . . . 99 ) × 128
| 5 | → | 51 |
| 15 | → | 51 |
| 25 | → | 52 |
| 35 | → | 51 |
| 45 | → | 51 |
| 55 | → | 51 |
| 65 | → | 51 |
| 75 | → | 52 |
| 85 | → | 51 |
| 95 | → | 51 |
| × | ||
| 128 | → | 27 |
512 + 27 will make zero but since 2 comes 7 times, so only 7 zero will come.
88. The largest among the numbers ?
a) $${\left( {0.1} \right)^2}$$
b) $$\sqrt {0.0121\,} $$
c) 0.12
d) $$\sqrt {0.0004} $$
Explanation:
| (0.1)2 | → | .01 | |
| $$\sqrt {0.0121} $$ | → | 0.11 | |
| 0.12 | → | 0.12 | → Largest |
| $$\sqrt {.0004} $$ | → | 0.02 |
89. The product of two numbers is 120 and the sum of their squares is 289. The sum of the two number is
a) 23
b) 7
c) 13
d) 169
Explanation: Let the two number are a and b
According to question,
$$\eqalign{ & {a^2} + {b^2} = 289 \cr & ab{\text{ }} = {\text{ }}120 \cr & {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \cr & {\left( {a + b} \right)^2} = {\left( {289} \right)^2} + \left( {2 \times 120} \right) \cr & {\left( {a + b} \right)^2} = 529 \cr & \left( {a + b} \right) = 23 \cr} $$
90. The digit in unit's place of the product (2153)167 is
a) 1
b) 3
c) 7
d) 9
Explanation:
$$\eqalign{ & {\left( {2153} \right)^{167}} \cr & {\text{unit digit = }}{{\text{3}}^{167}} \cr & {\text{unit digit }} \cr & {{\text{3}}^1} \to {\text{3}} \to {\text{3}} \cr & {{\text{3}}^2} \to 9 \to 9 \cr & {{\text{3}}^3} \to 27 \to 7 \cr & {{\text{3}}^4} \to 81 \to 1 \cr} $$
This cycle will continue
⇒ divide the power of 3 by 4
$$\eqalign{ & \frac{{167}}{4} \Rightarrow {\text{remainder is 3}} \cr & {{\text{3}}^3} \Rightarrow 7 \cr & {\text{unit digit = }}1 \times 7 = {\text{ }}7 \cr} $$
91. If a and b are two distinct natural numbers, which one of the following is true ?
a) $$\sqrt {a + b} > \sqrt a + \sqrt b $$
b) $$\sqrt {a + b} = \sqrt a + \sqrt b $$
c) $$\sqrt {a + b} < \sqrt a + \sqrt b $$
d) $$ab = 1$$
Explanation:
$$\eqalign{ & \sqrt {a + b\,} \,{\text{and }}\sqrt a + \sqrt b \cr & {\text{Squaring both sides}} \cr & {(\sqrt {a + b\,} )^2}{\text{and (}}\sqrt a + \sqrt b {)^2} \cr & \Rightarrow a + b\,\,{\text{and}}\,a + b + 2\sqrt a \sqrt b \cr & So\,\sqrt {a + b\,} \, < {\text{ }}\sqrt a + \sqrt b \cr} $$
92. Which of the following fraction is the smallest ?
$$\frac{7}{6},\,\frac{7}{9},\,\frac{4}{5},\,\frac{5}{7}$$
a) $$\frac{7}{6}$$
b) $$\frac{7}{9}$$
c) $$\frac{4}{5}$$
d) $$\frac{5}{7}$$
Explanation:
$$\frac{7}{6},\,\frac{7}{9},\,\frac{4}{5},\,\frac{5}{7}$$
Step 1 : Compare two fractions
$$\frac{{7}}{{6}}$$ ⤩ $$\frac{{7}}{{9}}$$
Cross multiply
63 , 42
↑ ↑
$$\frac{7}{6}$$ ⤩ $$\frac{7}{9}$$
42 is smaller than 63
So, $$\frac{7}{9}$$ is smaller than $$\frac{7}{6}$$
Step 2 : Compare
(i) $$\frac{7}{9},\frac{4}{5}$$
(ii) $$\frac{7}{9}$$ ⤩ $$\frac{4}{5}$$
Cross multiply it
35 , 36
↑ ↑
$$\frac{7}{9}$$ ⤩ $$\frac{4}{5}$$
35 is smaller than 36
So, $$\frac{7}{9}$$ is smaller than $$\frac{4}{5}$$
Step 3 : Compare
(i) $$\frac{7}{9},\frac{5}{7}$$
(ii) $$\frac{7}{9}$$ ⤩ $$\frac{5}{7}$$
Cross multiply it
49 , 45
↑ ↑
$$\frac{7}{9}$$ ⤩ $$\frac{5}{7}$$
45 is smaller than 49
S0, $$\frac{5}{7}$$ is the smallest
93. $$\frac{1}{5}$$ of a number exceeds $$\frac{1}{7}$$ of the same number by 10. The number is
a) 25
b) 150
c) 175
d) 200
Explanation: Let, the number be 'x'
According to the question
$$\eqalign{ & \frac{1}{5}x - \frac{1}{7}x = 10 \cr & \frac{{7x - 5x}}{{35}} = 10 \cr & 2x = 350 \cr & x = \frac{{350}}{2} \cr & x = 175 \cr} $$
94. The product of two numbers is 0.008. One of the number is $$\frac{1}{5}$$ of the other. The smaller number is
a) 0.2
b) 0.4
c) 0.02
d) 0.04
Explanation:
$$\eqalign{ & {\text{Let the one number be }}x \cr & {\text{And the other number be }}\frac{1}{5}x \cr & {\text{According to question}} \cr & x \times \frac{1}{5}x = 0.008 \cr & {x^2} = 0.040 \cr & {\text{ }}x = 0.2 \cr & {\text{Other number}} \cr & = \frac{1}{5} \times 0.2\, \cr & = 0.04 \cr & {\text{Smaller number is 0}}{\text{.04}} \cr} $$
95. When an integer K is divided by 3 the remainder is 1, and when K + 1 is divided by 5, The remainder is '0'. Of the following the possible value of K is
a) 62
b) 63
c) 64
d) 65
Explanation: Always do these types of question by option to save time
Pick up the option and follow the the question instruction
Take option 3
64 ⇒ Divided 3 it gives remainder 1
Now add 1 to 64
$$\frac{{65}}{5}$$ ⇒ remainder '0' is satisfies
So K = 64 this is answer
96. The sum of even numbers between 1 and 31 is:
a) 6
b) 28
c) 240
d) 512
Explanation: Let Sn = (2 + 4 + 6 + ..... + 30). This is an A.P. in which a = 2, d = 2 and l = 30
Let the number of terms be n Then,
a + (n - 1)d = 30
⇒ 2 + (n - 1) x 2 = 30
⇒ n = 15
$$\eqalign{ & \therefore {S_n} = \frac{n}{2}\left( {a + I} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{15}}{2} \times \left( {2 + 30} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 15 \times 16 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 240 \cr} $$
97. If (64)2 - (36)2 = 20 × x, then x = ?
a) 70
b) 120
c) 180
d) 140
Explanation:
$$\eqalign{ & 20 \times x = \left( {64 + 36} \right)\left( {64 - 36} \right) \cr & 20 \times x = 100 \times 28 \cr & \therefore {\kern 1pt} x = \frac{{100 \times 28}}{{20}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 140 \cr} $$
98. What is the unit digit in the product (365 × 659 × 771)?
a) 1
b) 2
c) 4
d) 6
Explanation: Unit digit in 34 = 1 ⇒ Unit digit in (34)16 = 1
Therefore Unit digit in 365 = Unit digit in [ (34)16 × 3 ] = (1 × 3) = 3
Unit digit in 659 = 6
Unit digit in 74 ⇒ Unit digit in (74)17 is 1
Unit digit in 771 = Unit digit in [(74)17 × 73] = (1 × 3) = 3
Therefore Required digit = Unit digit in (3 × 6 × 3) = 4
99. A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. When it is successively divided by 5 and 4, then the respective remainders will be
a) 1, 2
b) 2, 3
c) 3, 2
d) 4, 1
Explanation: 4 | x y = (5 × 1 + 4) = 9
------------
5 | y - 1 x = (4 × y + 1) = (4 × 9 + 1) = 37
------------
| 1 - 4
Now, 37 when divided successively by 5 and 4, we get
5 | 37
-------------
4 | 7 - 2
-------------
| 1 - 3
Respective remainders are 2 and 3
100. The sum of all three digit numbers, each of which on divide by 5 leaves remainder 3, is
a) 180
b) 1550
c) 6995
d) 99090
Explanation:
$$\eqalign{ & {\bf{Series :}} \cr & 103 + 108\, + ........ + \,998 \cr & a{\text{ }} = {\text{ }}103 \cr & d{\text{ }} = {\text{ }}5 \cr & {\text{Last term = 998}} \cr & {\text{Number of term }} \cr & = \frac{{998 - 103}}{5} + 1 \cr & = \frac{{895}}{5} + 1 \cr & = 180 \cr & {\text{Sum of n terms}} \cr & = \frac{n}{2}\left[ {2a{\text{ }} + \left( {n - 1} \right)d} \right] \cr & = \frac{{180}}{2}\left[ {2 \times 103 + \left( {180 - 1} \right)5} \right] \cr & = 99090 \cr} $$