Practice | Hashtable

21. The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:

A. 15 cm

B. 25 cm

C. 35 cm

D. 42 cm

Solution:

Required length = H.C.F. of 700 cm, 385 cm and 1295 cm
= 35 cm

22. Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:

A. 75

B. 81

C. 85

D. 89

Discuss

Solution:

Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29

\begin{aligned} First number \\ = \frac{551}{29} \\ = 19 \\ Third number \\ = \frac{1073}{29} \\ = 37 \\ ∴ Required sum \\ = 19 + 29 + 37 \\ = 85 \end{aligned}

23. Find the highest common factor of 36 and 84

A. 4

B. 6

C. 12

D. 18

Discuss

Solution:

36 = 2² × 3²
84 = 2² × 3 x 7
∴ H.C.F. = 2² × 3 = 12

24. Which of the following fraction is the largest ?

A. $\frac{7}{8}$

B. $\frac{13}{16}$

C. $\frac{31}{40}$

D. $\frac{63}{80}$

Discuss

Solution:

\begin{aligned} L .C .M . of 8, 16, 40 and 80 = 80 \\ \frac{7}{8} = \frac{70}{80}; \frac{13}{16} = \frac{65}{80}; \frac{31}{40} = \frac{62}{80} \\ Since, \frac{70}{80} > \frac{65}{80} > \frac{63}{80} > \frac{62}{80}, \\ So, \frac{7}{8} > \frac{13}{16} > \frac{63}{80} > \frac{31}{40} \\ So, \frac{7}{8} is the largest . \end{aligned}

25. The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:

A. 504

B. 536

C. 544

D. 548

Discuss

Solution:

Required number
= (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548

26. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

A. 123

B. 127

C. 235

D. 305

Discuss

Solution:

Required number
= H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127

27. Which of the following has the most number of divisors?

A. 99

B. 101

C. 176

D. 182

Discuss

Solution:

99 = 1 × 3 × 3 × 11
101 = 1 × 101
176 = 1 × 2 × 2 × 2 × 2 × 11
182 = 1 × 2 × 7 × 13
So, divisors of 99 are 1, 3, 9, 11, 33, 99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182
Hence, 176 has the most number of divisors.

28. The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:

A. 28

B. 32

C. 40

D. 64

Discuss

Solution:

Let the numbers be 2x and 3x
Then, their L.C.M. = 6x
So, 6x = 48 or x = 8
∴ The numbers are 16 and 24
Hence, required sum = (16 + 24) = 40

29. The H.C.F. of $\frac{9}{10},$ $\frac{12}{25},$ $\frac{18}{35}$ and $\frac{21}{40}$ is:

A. $\frac{3}{5}$

B. $\frac{252}{5}$

C. $\frac{3}{1400}$

D. $\frac{63}{700}$

Discuss

Solution:

\begin{aligned} Required H .C .F . \\ = \frac{H .C .F . of 9, 12, 18, 21}{L .C .M . of 10, 25, 35, 40} \\ = \frac{3}{1400} \end{aligned}

30. If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

A. $\frac{55}{601}$

B. $\frac{601}{55}$

C. $\frac{11}{120}$

D. $\frac{120}{11}$

Discuss

Solution:

\begin{aligned} Let the numbers be a and b \\ Then, a + b = 55 and \\ a b = 5 \times 120 = 600 \\ ∴ The required sum \\ = \frac{1}{a} + \frac{1}{b} = \frac{a + b}{a b} = \frac{55}{600} = \frac{11}{120} \end{aligned}