## Solved Cowbell maths quiz past questions and answers

### By Naija Scholarships Buzz

The following are sample solved questions from Cowbell mathematics quiz. It is important to note that CALCULATORS are not allowed through out the entire quiz!

#### — JUNIOR

Q1. A bottle of Cowbell milk is emptied into eight cups, each of which holds 150ml.
How many litters of milk was in the bottle?

#### Ans:  $\inline&space;\dpi{100}&space;\bg_white&space;\fn_jvn&space;\huge&space;1&space;\thinspace&space;litre&space;=&space;10^{-3}&space;milliliters.&space;\thinspace&space;\thinspace&space;Then&space;\thinspace&space;150ml&space;=\frac{150}{1000}l&space;=&space;\frac{3}{20}l.&space;\thinspace&space;Now&space;\thinspace&space;there&space;\thinspace&space;are&space;\thinspace&space;\frac{3}{20}l&space;\thinspace&space;in&space;\thinspace&space;each&space;\thinspace&space;cup,&space;\thinspace&space;so&space;\thinspace&space;that&space;\thinspace&space;8\times\frac{3}{20}l&space;=&space;\frac{6}{5}l&space;=&space;1.2&space;\thinspace&space;litres&space;\thinspace&space;is&space;\thinspace&space;the&space;\thinspace&space;required&space;\thinspace&space;quantity&space;\thinspace&space;in&space;\thinspace&space;the&space;\thinspace&space;bottle&space;\thinspace&space;of&space;\thinspace&space;Cowbell&space;\thinspace&space;milk.$

Q2. Eight people share a prize and receive 125 each. How much would each have received if there had been only five prize winners?

Ans:  Since eight people received 125 each, then the total money shared was 8 ×  125 = ₦1000. Now, if five people share  ₦1000, they would each get  ₦200 which is the required answer.

Q3. Mr. Dauda invested  ₦10,800 and at the end of each year he withdrew the interest. After 4 years he had withdrawn a total of  ₦3,240 in interest. At what annual rate of interest was his money invested?

Ans: Let I = (PRT/100), where I = Interest, P = Principal, and T = Time. Now, R = (100 × I)/(PT) = (100
×3240)/(10800 × 4) = 7(1/2)%.

Q4. John drives along a road which has two sets of traffic lights. The probability of their being green are 3/4 and 2/5 respectively.  What is the probability of John finding both sets not green?

Ans: The probability of both sets not green is (1 - 3/4) × (1 - 2/5) = 3/20.

Q5. A rectangle has a length which is double its width. The perimeter of the rectangle is 18cm. Find the width of the rectangle.

Ans: Let L = length, W = width. Then perimeter = 2L + 2W = 2(2W) + 2W = 6W, since L = 2W. Therefore, W 18/6 = 3cm which is the required width.

#### — SENIOR

Q1. Given that X varies inversely as the cube of Y and that X = 240 when Y = 3. Find X when Y = 2.

Ans: From the question, X = K/(Y^3). So when X = 240, Y = 3, we have K = 6480.
Now, when Y = 2, X = 6480/(2^3) = 810 which is the required answer.

Q2. Calculate the area of a triangle whose lengths are 6cm, 8cm, and 10cm respectively.

Ans: Since 10cm is the longest of the lengths, the hypotenuse must be 10cm long. So the area is then, A = (1/2) × 8 × 6 = (1/2) × 6 × 8 = 24cm squared.

Q3. What is the standard deviation correct to 2 decimal places of a distribution whose variance is 6.27?

Ans: Let S be the standard deviation and V the variance. Then, S = √(V) = √(6.27) = 2.50 to 2 decimal places.

Q4. Mrs. Okon drives 5km north from his house and then 10km on a bearing of 060° to a market.  How far is the market to her home to the nearest km?

Ans:

Let X be the distance, then by the cosine rule, (X^2) = 25 +100 - 2×5×10×Cos120° = 175. So X = 13km to the nearest km.

Q5. Three Cowbell depots P, Q, and R are such that P to Q is 50km, P to R is 90km. The bearing of Q from P is 075° and the bearing of R from P is 310°. Find the distance between Q and R.

Ans:

Let the distance be RQ, then using cosine rule, RQ = √(2500 + 8100 - 9000Cos215°) = 134km to the nearest km.