Directions(1-5): LCM & HCF
1. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

A. 123 B. 127 C. 235 D. 305

Answer : Option B
Explanation: Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032 = 127.
2. Which of the following has the most number of divisors?

A. 99 B. 101 C. 176 D. 182

Answer : Option C
Explanation: 99 = 1 x 3 x 3 x 11 101 = 1 x 101 176 = 1 x 2 x 2 x 2 x 2 x 11 182 = 1 x 2 x 7 x 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.
3. 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

Answer: Option C
Explanation: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.
4. The H.C.F. of [latex]\frac{9}{10}[/latex], [latex]\frac{12}{25}[/latex], [latex]\frac{18}{35}[/latex] and [latex]\frac{21}{40}[/latex] is

A. [latex]\frac{3}{5}[/latex] B. [latex]\frac{252}{5}[/latex] C. [latex]\frac{3}{1400}[/latex] D. [latex]\frac{63}{700}[/latex]

Answer: Option C
Explanation: Required H.C.F. = [latex]\frac{H.C.F. of 9, 12, 18, 21}{L.C.M. of 10, 25, 35, 40 }[/latex] = [latex]\frac{3}{1400 }[/latex]
5. 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. [latex]\frac{55}{101}[/latex] B. [latex]\frac{601}{55}[/latex] C. [latex]\frac{11}{120}[/latex] D. [latex]\frac{120}{11}[/latex]

Answer: Option C
Explanation: Let the numbers be a and b. Then, a + b = 55 and ab = 5 x 120 = 600. ∴ The required sum = [latex]\frac{1}{a }[/latex] + [latex]\frac{1}{b}[/latex] = [latex]\frac{a + b}{ab}[/latex] = [latex]\frac{55}{600}[/latex] = [latex]\frac{11}{120}[/latex]

Directions(1-5): Simple Interest
1. A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is:

A. 5% B. 8% C. 12% D. 15%

Answer : Option C
Explanation: S.I. for 3 years = Rs. (12005 - 9800) = Rs. 2205. S.I. for 5 years = Rs. [[latex]\frac{2205}{3}[/latex] x 5] = Rs. 3675 Principal = Rs. (9800 - 3675) = Rs. 6125. ∴ (6000 + 6000*r*[latex]\frac{3}{12}[/latex]*100) = 1020*6 + [latex]\frac{1020r}{12}[/latex]*100(1 + 2 + 3 + 4 + 5) Hence, rate = [latex]\frac{100 × 3675}{6125 × 5}[/latex] %= 12%
2. Simple interest on a certain sum for 6 years is 9/25 of the sum. The rate of interest is ?

A. 6% B. 6 [latex]\frac{1}{2}[/latex]% C. 8% D. 8 [latex]\frac{1}{2}[/latex]%

Answer : Option A
Explanation: Rate = [latex]\frac{(SI ×100)}{(Principal ×Time)}[/latex] [latex]\frac{9}{25}[/latex] x [latex]\frac{100}{6}[/latex] = 6% per annum
3. A Principal amount give Rs.12750 as SI for 7 years at the rate of 10% to get the same amount of interest of 10 years, what will be the rate of interest ?

A. 5% B. 7% C. 9% D. 4%

Answer : Option B
Explanation: LSI = [latex]\frac{PNR}{100}[/latex] = [latex]\frac{12750*7*10}{100}[/latex] = 8925 SI = 8925 R = [latex]\frac{8925*100}{12750*10}[/latex] = 7
4. Vithya gave Rs.20000/- to Priya for 5 years and Rs.30000/- to Vijay for 2 years. If the rate of interests is the same and she received Rs.12800- as SI from both. What is the rate of interest?

A. 8% B. 7% C. 9% D. 11%

Answer : Option D
Explanation: [latex]\frac{20000*5*R}{12750*10}[/latex] + [latex]\frac{30000*2*R}{12750*10}[/latex] = 12800 1600R = 12800 R = [latex]\frac{12800}{1600 }[/latex] = 8
5. A man took a certain sum as loan from bank at a rate of 9% SI per annum. The man lends the same amount to another person at 15% SI per annum. If at the end of 4 years man made a profit of Rs. 9000 from the deal. What was the original sum ?

A. Rs.32,700 B. Rs.39000 C. Rs.37500 D. Rs.32500

Answer : Option C
Explanation: [latex]\frac{P*15*4}{100 }[/latex] – [latex]\frac{P*9*4}{100 }[/latex] = 9000 60P – [latex]\frac{36P}{100 }[/latex] = 9000 24P = 900000 P = 37500

Directions(1-5): Mensuration Problems
1. A cylindrical cistern whose diameter is 14 cm is partly filled with water. If a rectangular block of iron 22 cm in length, 18 cm in breadth and 7 cm in thickness is wholly immersed in water, by how many cm will the water level rise?

A. 10 cm B. 18 cm C. 12 cm D. 16 cm

Answer : Option B
Explanation: Volume of the Rectangular Block = 22 * 18 * 7 Radius of the Cistern = 7 cm Area of the Cylinder = Π * r² * h = [latex]\frac{22}{7}[/latex] * 7 * 7 * h [latex]\frac{22}{7}[/latex] * 7 * 7 * h = 22 * 18 * 7; h = 18 cm
2. The perimeter of a rectangle and a square is 80 cm each. If the difference between their areas is 100 cm. Find the sides of the rectangle.

A. 30cm, 10cm B. 40cm, 15cm C. 25cm, 10cm D. 20cm, 30cm

Answer : Option A
Explanation: 2(l + b) = 4a = 80 l + b = 40; a = 20 => [latex]{a}^{2}[/latex] = 400 [latex]{a}^{2}[/latex] – lb = 100; 400 – lb =100; lb = 300 [latex]{(l – b)}^{2}[/latex] = [latex]{(l + b)}^{2}[/latex] – 4lb [latex]{(l – b)}^{2}[/latex] = 1600 – 1200; l – b = 20;
3. A circular path runs round a circular garden. If the difference between the circumference of the outer circle and inner circle is 88m. Find the width of Path?

A. 14 m B. 15 m C. 18 m D. 13 m

Answer : Option A
Explanation: Width of the Road = R – r 2πR – 2πr = 88 R – r = 14 m
4. What is the radius of the circle whose area is equal to the sum of the areas of two circles whose radii are 20 cm and 21 cm?

A. 27m B. 28m C. 29m D. 25m

Answer : Option C
Explanation: π[latex]{R}^{2}[/latex] = πr[latex]{1}^{2}[/latex] + πr[latex]{2}^{2}[/latex] πR² = π(r[latex]{1}^{2}[/latex] + r[latex]{2}^{2}[/latex]) [latex]{R}^{2}[/latex] = (400 + 441) R = 29
5. A hollow cylindrical tube open at both ends is made of plastic 4 cm thick. If the external diameter be 54 cm and the length of the tube be 490 cm, find the volume of plastic.

A. 320000 c[latex]{m}^{3}[/latex] B. 340000 c[latex]{m}^{3}[/latex] C. 306300 c[latex]{m}^{3}[/latex] D. 308000 c[latex]{m}^{3}[/latex]

Answer : Option B
Explanation: External Radius = 27; Internal Radius = 23 Volume of Plastic = πh([latex]{R}^{2[/latex] – [latex]{r}^{2}[/latex]) = [latex]\frac{22}{7}[/latex] * 490([latex]{27}^{2}[/latex] – [latex]{23}^{2}[/latex]) = 308000 c[latex]{m}^{3}[/latex]