Home / iGCSE Mathematics (0580) : C1.4 Use directed numbers in practical situations. iGCSE Style Questions Paper 1

iGCSE Mathematics (0580) : C1.4 Use directed numbers in practical situations. iGCSE Style Questions Paper 1

Question

Idris has c toy cars.
Fadl has twice as many cars as Idris.
Baasim has three more cars than Fadl.
(a) Write down an expression, in terms of c, to complete each statement.
Fadl has ………………………………… cars.
Baasim has ………………………………… cars.

(b) Write down an expression, in terms of c, for the total number of cars the three children have.
Give your answer in its simplest form.

▶️Answer/Explanation

(a) Fadl has 2c cars.
Since Fadl has twice as many cars as Idris, we can express the number of cars Fadl has in terms of c, the number of cars Idris has. Therefore, Fadl has 2c cars.
Baasim has 2c + 3 cars.
Since Baasim has three more cars than Fadl, we can express the number of cars Baasim has in terms of the number of cars Fadl has.
Therefore,Fadl has 2c cars.
Baasim has 2c+3 cars.

(b) Idris has c toy cars.
Baasim has 2c + 3 cars.
Fadl has 2c cars.
The total number of cars the three children have is,
c + 2c + (2c + 3) = 5c + 3

Question

 In each part, fill in the missing number to make a correct statement.
(a) \((-6+11) \times \) ………………..= -20
(b) \(\frac{7}{8} = \frac{……………}{176}\)

▶️Answer/Explanation

(a)To solve for the missing number, we can simplify the expression inside the parentheses first
\((-6+11) \times x = -20\)
\(5 \times x = -20\)
\(x=-4\)
\(\therefore \) \((-6+11) \times \)  -4 = -20
(b)To solve for the missing number, we can cross-multiply and solve for the unknown numerator,
\(\frac{7}{8} = \frac{x}{176}\)
\(7\times 176=8\times x\)
8x=1232
x=154
\(\therefore\) \( \frac{7}{8}=\frac{154}{176}\)

Question

 Without using your calculator, and leaving your answer as a fraction, work out
2\(\frac{1}{6}-\frac{7}{12}\)
You must show all your working.

▶️Answer/Explanation

1\(\frac{7}{12} or \frac{9}{12}\)

Firstly, we solve the mixed equation as follows:
2\(\frac{1}{6} – \frac{7}{12}\) = (2 × 6 + 1) / 6 = 13/6
Now using this evaluation , we can solve further as follows:
=13/6 – 7/12
=(13/6) \(\times\) (2/2) – (7/12)
=26/12 – 7/12
=19/12

Question

Piet, Rob and Sam collect model aeroplanes.
Piet has x aeroplanes.
Rob has 7 more aeroplanes than Piet.
Sam has three times as many aeroplanes as Piet.
(a) Write down an expression, in terms of x, for
(i) the number of aeroplanes Rob has,

(ii) the number of aeroplanes Sam has.

(b) The total number of aeroplanes is 32.
(i) Use the information in part (a) to write down an equation in x.

(ii) Solve your equation.

Write down the number of aeroplanes Rob has.

▶️Answer/Explanation

(a) (i) x + 7
(ii) 3x
(b) (i) x+their (a)(i)+their (a)(ii)=32
or better
(ii) (x =) 5

(c) 12

(a)Let’s say Piet has x be the number of aero planes.

(i) Rob has7 more aero planes than Piet So,
Number of aero planes Rob have = x + 7

(ii) It is given that Sam has three times as many aero planes as Piet So,
Number of aero plane Sam have = 3x

Hence, the expressions are as follows:
Number of aero planes Rob’s have = x + 7
Number of aero planes Sam have = 3x

b(i) Given : Total Number of aero planes = 32

Number of Piet’s aero planes + Number of Rob’s aero planes + Number of Sam’s aero planes = 32
Hence , Equation is : x + (x +7) + 3\(\times\) x = 32

(ii) Solving the above equation :
x + (x +7) + 3\(\times\) x = 32
5 \(\times\) x + 7 = 32
5 \(\times\) x = 32 – 7
5 \(\times\) x = 25
x = 5

Number of aero planes Piet have = x = 5
Number of aero planes Rob have = x + 7 = 5 + 7 = 12
Number of aero planes Sam have = 3 \(\times\) x = 3 \(\times\) 5 = 15

(c) Number of aero planes Sam have = 3 \(\times\) x = 3 \(\times\) 5 = 15

Scroll to Top