Practice | Hashtable

91. The HCF and LCM of 2⁴, 8², 16², 20³ are :

A. 2³, 32000

B. 2⁴, 32000

C. 2⁴, 25600

D. 2², 3200

Solution:

HCF of 2⁴, 8², 16², 20³ = 2⁴
LCM of 2⁴, 8², 16², 20³ = 2⁴ = 2⁸ × 125 = 32000

92. 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

Discuss

Solution:

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

93. 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

Discuss

Solution:

Let one part of string is x.
Now,
x +

\frac{9 x}{4}

= 221 m
x = 68 meter
and,

\frac{9 x}{4}

= 153 m
Difference between two parts = 153 - 68 = 85m.

94. The sum of 100 terms of the series 1 - 3 + 5 - 7 + 9 - 11 .......... is:

A. 100

B. - 200

C. 200

D. - 100

Discuss

Solution:

1 - 3 + 5 - 7 + 9 - 11 .......... + 197 - 199
= (- 2) + (-2) + (-2) + .......... + (-2) (50 times)
= 50 × (-2) = -100

95. The remainder of $\frac{6^{36}}{215} :$

A. 0

B. 1

C. 2

D. 3

Discuss

Solution:

\begin{aligned} \frac{6^{36}}{215}, can be written as \\ \frac{{(6^{3})}^{12}}{215} \\ O r, \frac{216^{12}}{215}, [216 on divided by 215, gives remainder 1] \\ \frac{1^{12}}{215} \\ The remainder will be 1 \end{aligned}

96. The remainder when (12¹³ + 23¹³) is divided by 11.

A. 0

B. 1

C. 2

D. 3

Discuss

Solution:

\begin{aligned} \frac{12^{13}}{11} gives Remainder 1 \\ It can be written as \\ \frac{(12 \times 12 \times 12 \times 12 . . . . . . . . 13 times)}{11} \\ On dividing it gives remainder 1, each time . \\ 1 \times 1 \times 1 \times 1 . . . . . . . . . 13 times \\ So, final remainder will be 1 \\ \frac{23^{13}}{11} \Rightarrow Remainder 1 \\ Thus, \\ The remainder of \frac{(12^{13} + 23^{13})}{11} \\ = (1 + 1) \\ = 2 \end{aligned}

97. The remainder when 75^75⁷⁵ is divide by 37.

A. 0

B. 1

C. 5

D. 7

Discuss

Solution:

\frac{75^{75^{75}}}{37}

When 75 is divided by 37, it leaves remainder of 1.
So, 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

98. 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

Discuss

Solution:

The only possible number which is both perfect square and perfect cube is 729 which is cube of 9 and square of 27
So, 7 × 2 × 9 = 126

99. 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

Discuss

Solution:

Going through options:
LCM of 204 and 170 = 1020.

100. 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

Discuss

Solution:

\begin{aligned} Total Marks = 90 \\ Marks in QA = \frac{7}{9} of 90 \\ = \frac{(7 \times 90)}{9} \\ = 7 \times 10 \\ = 70 \end{aligned}