**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)

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

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

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)

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)

A. 100π

B. 169π

C. 69π✔️

D. 9π

E. 269π

**SOLUTION**

Area between two concentric circles is πR² – πr²

=π(13² – 10²)

= 69π (C)

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)

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)

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**

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