Mathematical And Numerical Reasoning 2026 Njfp Job
Mathematical And Numerical Reasoning 2026 Njfp Job
Question 21
What is 999 times 100.0?
To multiply a whole number by 100, you simply shift the decimal point two places to the right or append two zeros to the end of the number. The decimal '.0' does not change the value of the multiplier. The calculation is as follows: \[ 999 \times 100 = 99,900 \] This follows the basic powers-of-ten multiplication rule.
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Question 22
A car can cover a distance of 540 km on 18 litres of petrol. How far can it travel on 16 litres of petrol?
To solve this, we first find the distance the car covers per litre (fuel efficiency):\[ \text{Fuel efficiency} = \frac{540 \text{ km}}{18 \text{ litres}} = 30 \text{ km/litre} \]Now, we calculate the distance for 16 litres:\[ \text{Distance} = 30 \text{ km/litre} \times 16 \text{ litres} = 480 \text{ km} \]Note: The provided answer B (223 km) is mathematically incorrect. The correct calculation yields 480 km.
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Question 23
4 workers get paid ₦160,000 for working for five days. How much will 5 workers get paid for a day?
First, find the total pay per day for the group of 4 workers:\[ \frac{160,000}{5 \text{ days}} = 32,000 \text{ per day for 4 workers} \]Next, find the pay for one worker per day:\[ \frac{32,000}{4 \text{ workers}} = 8,000 \text{ per worker per day} \]Finally, calculate the pay for 5 workers for one day:\[ 8,000 \times 5 = 40,000 \]
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Question 24
The speed of a car is 80 km/hr. Distance covered in 1 1/4 hours is ________.
Distance is calculated by multiplying speed by time. First, convert the mixed fraction time to a decimal or improper fraction:\[ 1 \frac{1}{4} \text{ hours} = 1.25 \text{ hours} \text{ or } \frac{5}{4} \text{ hours} \]Using the formula:\[ \text{Distance} = \text{Speed} \times \text{Time} \]\[ \text{Distance} = 80 \text{ km/hr} \times 1.25 \text{ hr} = 100 \text{ km} \]
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Question 25
A shopkeeper purchased 200 bulbs for ₦10 each. However, 5 bulbs were fused and had to be thrown away. The remaining were sold at ₦12 each. What will be the percentage profit?
To find the percentage profit, we follow these steps:
1. Calculate Total Cost Price (CP):\[ CP = 200 \text{ bulbs} \times \text{₦}10 = \text{₦}2,000 \]2. Calculate the number of bulbs sold:\[ 200 - 5 = 195 \text{ bulbs} \]3. Calculate Total Selling Price (SP):\[ SP = 195 \text{ bulbs} \times \text{₦}12 = \text{₦}2,340 \]4. Calculate the Profit:\[ \text{Profit} = SP - CP = 2,340 - 2,000 = \text{₦}340 \]5. Calculate Profit Percentage:\[ \text{Profit \%} = \left( \frac{\text{Profit}}{CP} \right) \times 100 \]\[ \text{Profit \%} = \left( \frac{340}{2000} \right) \times 100 = 17\% \]
1. Calculate Total Cost Price (CP):\[ CP = 200 \text{ bulbs} \times \text{₦}10 = \text{₦}2,000 \]2. Calculate the number of bulbs sold:\[ 200 - 5 = 195 \text{ bulbs} \]3. Calculate Total Selling Price (SP):\[ SP = 195 \text{ bulbs} \times \text{₦}12 = \text{₦}2,340 \]4. Calculate the Profit:\[ \text{Profit} = SP - CP = 2,340 - 2,000 = \text{₦}340 \]5. Calculate Profit Percentage:\[ \text{Profit \%} = \left( \frac{\text{Profit}}{CP} \right) \times 100 \]\[ \text{Profit \%} = \left( \frac{340}{2000} \right) \times 100 = 17\% \]
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Question 26
By selling a water heater for ₦3,200, the shopkeeper incurs a loss of 20%. If he wants to earn a profit of 20%, at what price should he sell the water heater?
Step 1: Find the Cost Price (CP). A 20% loss means the selling price is 80% of the CP:\[ 0.80 \times CP = 3,200 \]\[ CP = \frac{3,200}{0.80} = \text{₦}4,000 \]Step 2: Find the target Selling Price (SP) for a 20% profit. This means the SP should be 120% of the CP:\[ \text{Target SP} = 1.20 \times 4,000 \]\[ \text{Target SP} = \text{₦}4,800 \]
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Question 27
The smallest number which when diminished by 3 is divisible by 21, 28, 36 and 45 is?
To find a number that is divisible by a set of numbers, we first find their Least Common Multiple (LCM).
Prime factorisation:
\( 21 = 3 \times 7 \)
\( 28 = 2^2 \times 7 \)
\( 36 = 2^2 \times 3^2 \)
\( 45 = 3^2 \times 5 \)
The LCM is the product of the highest power of each prime factor:\[ LCM = 2^2 \times 3^2 \times 5 \times 7 = 4 \times 9 \times 5 \times 7 = 1,260 \]The problem states that the required number, when 'diminished by 3', becomes divisible by these numbers. This means our number \( x \) satisfies:\[ x - 3 = 1,260 \]\[ x = 1,260 + 3 = 1,263 \]
Prime factorisation:
\( 21 = 3 \times 7 \)
\( 28 = 2^2 \times 7 \)
\( 36 = 2^2 \times 3^2 \)
\( 45 = 3^2 \times 5 \)
The LCM is the product of the highest power of each prime factor:\[ LCM = 2^2 \times 3^2 \times 5 \times 7 = 4 \times 9 \times 5 \times 7 = 1,260 \]The problem states that the required number, when 'diminished by 3', becomes divisible by these numbers. This means our number \( x \) satisfies:\[ x - 3 = 1,260 \]\[ x = 1,260 + 3 = 1,263 \]
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Question 28
A cone, hemisphere and a cylinder stand on the same base and have the same height. The ratio of the volumes is?
Let the radius of the base be \( r \). Since they have the same height and stand on the same base as the hemisphere, the height \( h \) must equal the radius \( r \).
1. Volume of Cone: \( V_1 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^3 \)
2. Volume of Hemisphere: \( V_2 = \frac{2}{3} \pi r^3 \)
3. Volume of Cylinder: \( V_3 = \pi r^2 h = \pi r^3 \)
Taking the ratio:\[ V_1 : V_2 : V_3 = \frac{1}{3}\pi r^3 : \frac{2}{3}\pi r^3 : \pi r^3 \]Dividing through by \( \pi r^3 \):\[ \frac{1}{3} : \frac{2}{3} : 1 \]Multiplying by 3 to clear the fractions:\[ 1 : 2 : 3 \]Note: The provided key indicated B, but the standard derivation for Cone:Hemisphere:Cylinder volumes results in 1:2:3.
1. Volume of Cone: \( V_1 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^3 \)
2. Volume of Hemisphere: \( V_2 = \frac{2}{3} \pi r^3 \)
3. Volume of Cylinder: \( V_3 = \pi r^2 h = \pi r^3 \)
Taking the ratio:\[ V_1 : V_2 : V_3 = \frac{1}{3}\pi r^3 : \frac{2}{3}\pi r^3 : \pi r^3 \]Dividing through by \( \pi r^3 \):\[ \frac{1}{3} : \frac{2}{3} : 1 \]Multiplying by 3 to clear the fractions:\[ 1 : 2 : 3 \]Note: The provided key indicated B, but the standard derivation for Cone:Hemisphere:Cylinder volumes results in 1:2:3.
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Question 29
In the following question a number series is given with one term missing. Choose the correct alternative that will continue the same pattern and fill in the blank spaces.
2, 1, (1/2), (1/4), ?
2, 1, (1/2), (1/4), ?
This is a geometric progression where each subsequent term is found by multiplying the previous term by a constant ratio (common ratio \( r \)).
Observe the terms:\[ 1 \div 2 = \frac{1}{2} \]\[ \frac{1}{2} \div 1 = \frac{1}{2} \]\[ \frac{1}{4} \div \frac{1}{2} = \frac{1}{4} \times 2 = \frac{1}{2} \]The common ratio is \( \frac{1}{2} \). To find the next term:\[ \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \]
Observe the terms:\[ 1 \div 2 = \frac{1}{2} \]\[ \frac{1}{2} \div 1 = \frac{1}{2} \]\[ \frac{1}{4} \div \frac{1}{2} = \frac{1}{4} \times 2 = \frac{1}{2} \]The common ratio is \( \frac{1}{2} \). To find the next term:\[ \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \]
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Question 30
A man invested ₦5000 for 9 months at 4 %. What is the simple interest?
To calculate the simple interest, we use the standard formula:\[ I = \frac{P \times R \times T}{100} \]Where:
\( P \) (Principal) = ₦5,000
\( R \) (Rate) = 4% per annum
\( T \) (Time) = 9 months. Since the rate is usually per annum, we must express time in years: \( \frac{9}{12} \) or \( 0.75 \) years.
Substituting the values into the formula:\[ I = \frac{5000 \times 4 \times 9}{100 \times 12} \]\[ I = \frac{180000}{1200} \]\[ I = 150 \]The total interest earned over the 9-month period is ₦150.
\( P \) (Principal) = ₦5,000
\( R \) (Rate) = 4% per annum
\( T \) (Time) = 9 months. Since the rate is usually per annum, we must express time in years: \( \frac{9}{12} \) or \( 0.75 \) years.
Substituting the values into the formula:\[ I = \frac{5000 \times 4 \times 9}{100 \times 12} \]\[ I = \frac{180000}{1200} \]\[ I = 150 \]The total interest earned over the 9-month period is ₦150.
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