Q.1.The ratio of the present ages of Priya and her mother is 3: 7. The mother’s age at the time of Priya’s birth was 48 years. Find the mother’s present age.

a) 84 yrs

b) 42 yrs

c) 36 yrs

d) None of these

Sol : Option A Explanation: Present ratio is 7 : 3.∴ actual ages are 7x and 3x. ∴ 7x – 3x = 48 → x = 12. Hence their present ages are 84 and 36 yrs.

Q.2. The perimeter of a rectangle is 64 cm. If the ratio of the lengths of two adjacent sides is 7: 9, find the lengths of these sides.

a) 24 cm, 28 cm

b) 14 cm, 18 cm

c) 7 cm, 9 cm

d) None of these

Sol : Option B Explanation: : Perimeter of a rectangle = 2 (L + B) Also L: B = 7 : 9 ∴ actual values are 5x and 8x. Hence 2(7x + 9x) = 64. → x = 2 ∴ sides will be of 14 cm and 18 cm.

Q.3. The ratio of the length and the breadth of a rectangle is 3 : 5 and its area is 1.35 cm^{2}. Find the length of the rectangle.

a) 9 cm

b) 0.9 cm

c) 0.09 cm

d) 90 cm

Sol : Option B Explanation: Area of a rectangle = LB. Ratio of sides = 3 : 5. ∴(3x)(5x) = 1.35 → x = 0.3. → Length of the rectangle = 3 (0.3) = 0.9 cm and breadth = 5(0.3) = 1.5 cm.

Q.4.The ratio of the present ages of John and Jim is 5 : 3. Four years hence it will be 3 : 2. Find the present age of John.

a) 5yrs

b) 5 yrs

c) 10 yrs

d) 20yrs

Sol : Option D Explanation: : Present ratio = 5 : 3. ∴ actual values are 5x and 3x. So (5x + 4) / (3x + 4) = 3/2 → x = 4 So present ages are 20 yrs and 12 yrs.

Q.5.There are 145 students in the first three standards. The ratio of number of students in the first and the second standards is 2 : 3, while that of students in standards second and third is 4: 3. Find the number of students in 2^{nd} standard.

a) 40 yrs

b) 60 yrs

c) 45 yrs

d) 65yrs

Sol : Option B Explanation: : Total students = 145. Ratio of students in 1st and 2nd standards = 2 : 3 Ratio of students in 2^{nd} and 3^{rd} standards = 4 : 3 Hence combined ratio i.e. 1^{st}: 2^{nd}: 3^{rd} is 8 : 12 : 9. ∴ Number of students in each standard = (145 × 8)/29 = 40, (145 × 12)/29 = 60 and (145 × 9)/29 = 45.

Q.6.Which number when added to each of the numbers 24, 32 and 42 would make the sums to be in continued proportion?

a) 4

b) 5

c) 6

d) 8

Sol : Option D Explanation: Let the number to be added is x. ∴(24+ x) / (32 + x) = (32 + x) / (42 + x). Solving for x, we get x = 8.

Q7.If 46 – x is the geometric mean between 56 – x and 38 – x, find the value of x.

a) 4

b) 5

c) 6

d) 8

Sol : Option C Explanation: (38 – x)(56 – x) = (46 – x)^{2}. Solving this equation, x = 6.

Q8.If 6m – n = 4m + 13n, find the value of 2m + n : 2m – 3n.

a) 14:11

b) 15:14

c) 15:11

d) None of these

Sol : Option B Explanation: 6m – n = 4m + 13n → m = 7n. ∴ Required ratio = (2m+ n : 2m – 3n) =15n : 11n → 15 : 11.

Q9.The ratio of the measures ∠A and ∠B of a triangle ABC is 3 : 2. The ratio of the measures of ∠B and ∠C is 4 : 5. Find the measure of largest angle of the triangle ABC.

a) 72°

b) 78°

c) 48°

d) 60°

Sol : Option A Explanation: : ∠ A : ∠ B = 3 : 2. ∠ B : ∠ C = 4 : 5. ∴ ∠ A : ∠ B : ∠ C = 6 : 4 : 5. ∴ actual values are 6x, 4x and 5x. So 6x + 4x + 5x = 180 → 15x = 180 → x = 12. So angles are 6(12) = 72, 4(12) = 48 and 5(12) = 60

Q10.A piece of string 70 cm in length was cut into pieces, the ratio of whose lengths was 3: 7. Find the length of longest piece.

a) 21 cm

b) 70 cm

c) 49 cm

d) 7cm

Sol : Option C Explanation: Total length = 70 cm. Ratio is 3:7 ∴ Length of longest piece is (70 × 7) / 10 = 49cm