# JAMB 2021 POSSIBLE MATHEMATICS QUESTIONS

1. Change 71base10 to base 8
A. 107 base 8 ✔️
B. 106 base 8
C. 71 base 8
D. 17 base 8

SOLUTION
8|71
8|8 REM 7
8|1 REM 0
8|0 REM 1
ANSWER = 107 base 8 (A)

2. If N225.00 yields N 27.00 in x years simple interest at the rate of 4% per annum , find X
A. 3✔️
B. 4
C. 12
D. 17

SOLUTION
Principal = N225.00
Interest = N 27.00
Year = X
Rate = 4%
S.I = PTR / 100
27 = 225 × 4 × X / 100
T (years) = 3yrs

3. Calculate the standard deviation of the following data :
7,8,9,10,11,12,13.
A. 2✔️
B. 4
C. 10
D. 11

SOLUTION
Firstly find X , hence X = ( £x / N )
Ex = 7+8+9+10+11+12+13 = 70
N = 7
.°. X = 70/7 = 10

x | x – x | (x – x )²
7| -3 | 9
8| -2 | 4
9| -1 | 1
10| 0 | 1
11| 1 | 1
12| 2 | 4
13| 3 | 9
S.D = ( √£(×-×)²/N ) Note square root applies for both numerator and denominator.
= ( √£d²/N ) = √28/7 = √ 4 = 2 (A)

4. If y = 3t³ + 2t² – 7t + 3 , find dy/dt at t = -1
A. -1
B. 1
C. -2 ✔️
D. 2

READ  2021 jamb syllabus for physics

SOLUTION
dy/dt = 9t² + 4t – 7
When t = -1
dy/dt = 9(-1)² + 4(-1) – 7
= 9 – 4 – 7 = -2 (C)

5. A man wishes to keep some money in a savings deposit at 25% compound interest so that after 3 years he can buy a car of #150,000 . How much does he need to deposit now ?
A. #76800.00✔️
B. #85714.28
C. # 96000.00
D. # 112000.50

SOLUTION
A =#150,000
r = 25%
n = 3
Remember A = P ( 1 + r/100)^n
150,000 = P ( 1 + 0.25 ) ³
150,000 = P (1.25)³
P = 150,000 / 1.25³ = #76800.00(A)

6. A trader bought goats for #4000 each. He sold them for #180,000 at a loss of 25%. How many goats did he buy ?
A. 50
B. 60✔️
C. 36
D. 45

SOLUTION
Loss % = c.p – s.p / c.p × 100%

25 = c.p – 180,000 / c.p × 100
25 = 100 ( c.p – 180,000 ) / c.p
25 = 100cp – 18,000,000 / cp
Cross multiply
25cp = 100cp – 18,000,000
18,000,000 = 100cp – 25cp
18,000,000 = 75cp
C.P = #240,000

READ  JAMB CBT simplified and made easy with the use of ONLY 9 KEYS

Number of goats = 240,000 / 4000 = 60 (B)

7. The range of the data K+2 , k-3 , k+4 , k-2, k-5 , k+3 , k-1 and k+6 is
A. 11✔️
B. 10
C. 8
D. 6

SOLUTION
RANGE = highest – lowest
( K + 6 ) – ( K – 5 )
K + 6 – k + 5 = 11 (A)

8. What is the area between two concentric circles of diameters 26cm and 20cm ?
A. 100π
B. 169π
C. 69π✔️
D. 9π
E. 269π

SOLUTION
Area between two concentric circles is πR² – πr²
π(13² – 10²) = 69π (C)

9. Find the sum to infinity of the following series 3 + 2 + 4/3 + 8/9 + 16/27 + ________
A. 1270
B. 190
C. 18
D. 9✔️

SOLUTION
3 + 2 + 4/3 + 8/9 + 16/27
a = 3 , r = 2/3 ,
S∞ = a / 1 – r
= 3/1-2/3 = 9(D)

10. The derivative of cosec x is
A. tan x cosec x
B. – cot x cosec x✔️
C. tan x sec x
D. – cot x sec x

11. If y = ( 1 + x )² , find dy/dx
A. -1
B. 2 + 2x✔️
C. 1 + 2x
D. 2x – 1

READ  JAMB COMPUTER BASED TEST TUTORIAL

SOLUTION
If y = ( 1 + x )²
dy/dx = 2( 1 + x )
= 2 + 2x (B)

12. Which mathematics Question paper type is given to you ?
A. Type A
B. Type B
C. Type C
D. Type D ✔️

13. At what rate will the interest on #400 increase to #24 in 3years reckoning in simple interest?
A. 3%
B. 2%✔️
C. 5%
D. 4%

SOLUTION
S.I = PTR / 100 where
P denotes principal = #400
T denotes time = 3years
R denotes Rate = ?
24 = 400×3×R / 100
24×100 = 400×3×R
R = 2% (B)

14. If x*y = x + y² , find the value of ( 2*3 )*5
A. 25
B. 11
C. 55
D. 36✔️

SOLUTION
(2*3)*5 = (2+3²)*5
= (2+9)*5 = 11*5
Hence 11*5 = 11+5² = 11+25 = 36(D)

15. Simplify ( √6 + 2 )² ( √6 – 2 )²
A. 2√6
B. 4√6
C. 8√6✔️
D. 16√6

SOLUTION
( √6 + 2 )² (√6 – 2 )²
(√6 + 2 + √6 – 2 ) ( √6 + 2 -√6 + 2 )
(2√6 ) (4) = 8√6 (C)

error: Content is protected !!