能力傾向測試 - 基礎算術示例

答案 - A

解釋

Here a = 5, d = 8 - 5 = 3, n = 16
Using formula Tn = a + (n - 1)d
T16 = 5 + (16 - 1) x 3
= 50

答案 - C

解釋

Here a = 4, d = 9 - 4 = 5
Using formula Tn = a + (n - 1)d
Tn = 4 + (n - 1) x 5 = 109 where 109 is the nth term.
=> 4 + 5n - 5 = 109
=> 5n = 109 + 1 
=> n = 110 / 5 
= 22 

答案 - D

解釋

Here a = 7, d = 13 - 7 = 6, Tn = 205
Using formula Tn = a + (n - 1)d
Tn = 7 + (n - 1) x 6 = 205 where 205 is the nth term.
=> 7 + 6n - 6 = 205
=> 6n = 205 - 1 
=> n = 204 / 6 
= 34 

答案 - A

解釋

Using formula Tn = a + (n - 1)d
T6 = a + (6 - 1)d = 12   ...(i)
T8 = a + (8 - 1)d = 22   ...(ii)
Substract (i) from (ii)
=> 2d = 10 
=> d = 5
Using (i)
a = 12 - 5d 
= 12 - 25
= -13 

答案 - B

解釋

Using formula Tn = a + (n - 1)d
T6 = a + (6 - 1)d = 12   ...(i)
T8 = a + (8 - 1)d = 22   ...(ii)
Substract (i) from (ii)
=> 2d = 10 
=> d = 5

答案 - C

解釋

Using formula Tn = a + (n - 1)d
T6 = a + (6 - 1)d = 12   ...(i)
T8 = a + (8 - 1)d = 22   ...(ii)
Substract (i) from (ii)
=> 2d = 10 
=> d = 5
Using (i)
a = 12 - 5d 
= 12 - 25
= -13 
∴ T16 = -13 + (16 - 1) x 5
= 75 - 13 
= 62 

答案 - D

解釋

Here a = 5, d = 9 - 5 = 4, n = 17
Using formula Sn = (n/2)[2a + (n - 1)d]
S17 = (17/2)[2 x 5 + (17 - 1) x 4]
= (17/2)(10 + 64)
= 17 x 74 / 2
= 629

答案 - A

解釋

Here a = 2, d = 5 - 2 = 3, Tn = 182
Using formula Tn = a + (n - 1)d
a + (n - 1)d = 182
=> 2 + (n - 1) x 3 = 182
=> 3n = 183
=> n = 61.
Using formula Sn = (n/2)[2a + (n - 1)d]
S61 = (61/2)[2 x 2 + (61 - 1) x 3]
= (61/2)(4 + 180)
= 61 x 184 / 2
= 5612

答案 - B

解釋

Let've numbers are a - d, a and a + d
Then a - d + a + a + d = 15
=> 3a = 15
=> a = 5
Now (a - d)a(a + d) = 80
=> (5 - d) x 5 x (5 + d) = 80
=> 25 - d2 = 16
=> d2 = 9
=> d = +3 or -3
∴ numbers are either 2, 5, 8 or 8, 5, 2.

答案 - B

解釋

Here a = 3, r = 6 / 3 = 2, T9 = ?
Using formula Tn = ar(n - 1)
T9 = 3 x 2(9 - 1) 
=3 x 28 
=3 x 256
=768

答案 - A

解釋

Using formula Tn = ar(n - 1)
T4 = ar(4 - 1) = 54   
=> ar3 = 54   ...(i)
T9 = ar(9 - 1) = 13122
=> ar8 = 13122   ...(ii)
Dividing (ii) by (i)
=> r5 = 13122 / 54 = 243 = (3)5
=> r = 3
Using (i)
a x 27 = 54
=> a = 2 

答案 - B

解釋

Using formula Tn = ar(n - 1)
T4 = ar(4 - 1) = 54   
=> ar3 = 54   ...(i)
T9 = ar(9 - 1) = 13122
=> ar8 = 13122   ...(ii)
Dividing (ii) by (i)
=> r5 = 13122 / 54 = 243 = (3)5
=> r = 3

答案 - C

解釋

Using formula Tn = ar(n - 1)
T4 = ar(4 - 1) = 54   
=> ar3 = 54   ...(i)
T9 = ar(9 - 1) = 13122
=> ar8 = 13122   ...(ii)
Dividing (ii) by (i)
=> r5 = 13122 / 54 = 243 = (3)5
=> r = 3
Using (i)
a x 27 = 54
=> a = 2 
∴ T6 = ar(6 - 1) = 2 x (3)5  
= 2 x 243
= 486

答案 - A

解釋

Let the numbers are y and 80 - y.
Then 3y = 5(80-y)
=> 8y = 400 
∴ y = 50
and second number = 80 - 50 = 30.

答案 - B

解釋

Let the number be y.
Then (y / 3) - (y / 5) = 16
=> 5y - 3y = 16 x 15 = 240
=> 2y = 240
∴ y = 120

答案 - C

解釋

Let the numbers be 3y , 3y + 3, 3y + 6
Now 3y + 3y + 3 + 3y + 6 = 90
=> 9y = 81
=> y = 9
=> largest number = 3y + 6 = 3 x 9 + 6 
= 33

答案 - D

解釋

Let the positive integer by y.
Then y2 - 15y = 16
=> y2 - 15y - 16 = 0
=> y2 - 16y + y - 16 = 0
=> y(y-16) + (y-16) = 0
=> (y+1)(y-16)= 0
∴ y = 16. as -1 is not a positive integer.

答案 - A

解釋

Let the positive integer by y.
Then 23y - 2y2 = 63
=> 23y - 2y2 - 63 = 0
=> 2y2 - 23y + 63 = 0
=> 2y2 - 14y - 9y + 63 = 0
=> 2y(y-7) - 9(y-7)= 0
=> (2y-9)(y-7)= 0
∴ y = 7. as 9/2 is not an integer.

答案 - B

解釋

Let've number as 3y, 2y and 5y.
Then 9y2 + 4y2 + 25y2 = 1862.
=> 38y2 = 1862
=> y2 = 1862 / 38 = 49
=> y = 7
∴ smallest number = 2y = 2 x 7 = 14.

答案 - C

解釋

Let the ten's digit is x and unit digit of number is y.
Then  x + y = 10   ...(i)
(10x + y) - (10y - x) = 54
=> 9x - 9y = 54
=> x - y = 6    ...(ii)
Adding (i) and (ii)
2x = 16
=> x = 8
Using (i)
y = 10 - x = 2
∴ number is 82.