JAMB 2022 LIKELY QUESTION AND ANSWER ON MATHEMATICS

JAMB 2022 LIKELY QUESTION AND ANSWER ON MATHEMATICS

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

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 N150,000 . How much does he need to deposit now ?
A. N76800.00✔️
B. N85714.28
C. N96000.00
D. N112000.50

hey 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 = N240,000

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

SOLUTION

If y = ( 1 + x )²

dy/dx = 2( 1 + x )

= 2 + 2x (B)

12. At what rate will the interest on N400 increase to N24 in 3years reckoning in simple interest?
A. 3%
B. 2%✔️
C. 5%
D. 4%

SOLUTION

S.I = PTR / 100
where P denotes principal = N400

T denotes time = 3years

R denotes Rate = ?

24 = 400×3×R / 100

24×100 = 400×3×R

R = 2% (B)

13. 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)

14. 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)

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

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 N150,000 . How much does he need to deposit now ?
A. N76800.00✔️
B. N85714.28
C. N96000.00
D. N112000.50

hey 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 = N240,000

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

SOLUTION

If y = ( 1 + x )²

dy/dx = 2( 1 + x )

= 2 + 2x (B)

12. At what rate will the interest on N400 increase to N24 in 3years reckoning in simple interest?
A. 3%
B. 2%✔️
C. 5%
D. 4%

SOLUTION

S.I = PTR / 100
where P denotes principal = N400

T denotes time = 3years

R denotes Rate = ?

24 = 400×3×R / 100

24×100 = 400×3×R

R = 2% (B)

13. 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)

14. 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)

 

EXAMCHOKE JAMB 2022 SYSTEM PACKAGE EXAMPLE

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

*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 N150,000 . How much does he need to deposit now ?
A. N76800.00✔️
B. N85714.28
C. N96000.00
D. N112000.50

hey 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 = N240,000

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

SOLUTION

If y = ( 1 + x )²

dy/dx = 2( 1 + x )

= 2 + 2x (B)

12. At what rate will the interest on N400 increase to N24 in 3years reckoning in simple interest?
A. 3%
B. 2%✔️
C. 5%
D. 4%

SOLUTION

S.I = PTR / 100
where P denotes principal = N400

T denotes time = 3years

R denotes Rate = ?

24 = 400×3×R / 100

24×100 = 400×3×R

R = 2% (B)

13. 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)

14. 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)

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

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 N150,000 . How much does he need to deposit now ?
A. N76800.00✔️
B. N85714.28
C. N96000.00
D. N112000.50

hey 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 = N240,000

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

SOLUTION

If y = ( 1 + x )²

dy/dx = 2( 1 + x )

= 2 + 2x (B)

12. At what rate will the interest on N400 increase to N24 in 3years reckoning in simple interest?
A. 3%
B. 2%✔️
C. 5%
D. 4%

SOLUTION

S.I = PTR / 100
where P denotes principal = N400

T denotes time = 3years

R denotes Rate = ?

24 = 400×3×R / 100

24×100 = 400×3×R

R = 2% (B)

13. 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)

14. 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)

BY EXAMCHOKE TEAM